NCERT Solutions for Class 11 Maths Chapter 9 - Sequences And Series

Exercise 9.1: This exercise introduces the concept of sequences and their types, including arithmetic sequences, geometric sequences, and harmonic sequences. Students will practice identifying the nth term of each type of sequence.

Exercise 9.2: In this exercise, students will learn about the Arithmetic Progression (AP) and its various properties, such as the nth term and the sum of n terms. They will also practice solving problems related to these concepts.

Exercise 9.3: This exercise focuses on Geometric Progression (GP) and its various properties. Students will learn about the nth term and the sum of n terms of a GP and how to apply these concepts in problem-solving.

Exercise 9.4: This exercise covers the concept of the sum of the first n terms of an AP, which is also known as an arithmetic series. Students will learn to derive the formula for the sum of an AP and solve problems related to it.

Miscellaneous Exercise: This exercise includes a mix of questions covering all the concepts taught in the chapter. Students will have to apply their knowledge of sequences and series to solve various problems and answer questions.

Access NCERT Solutions for Class 11 Maths Chapter 9 – Sequences and Series

1. Write the first five terms of the sequences whose \[{{n}^{th}}\] term is \[{{a}_{n}}=n\left( n+2 \right)\] .

Ans:

The given equation is \[{{a}_{n}}=n\left( n+2 \right)\] .

Substitute \[n=1\] in the equation.

\[{{a}_{1}}=1\left( 1+2 \right)\]

\[\Rightarrow {{a}_{1}}=3\]

Similarly substitute \[n=2,3,4\] and \[5\]in the equation.

\[{{a}_{2}}=2\left( 2+2 \right)\]

\[\Rightarrow {{a}_{2}}=8\]

\[{{a}_{3}}=3\left( 3+2 \right)\]

\[\Rightarrow {{a}_{3}}=15\]

\[{{a}_{4}}=4\left( 4+2 \right)\]

\[\Rightarrow {{a}_{4}}=24\]

\[{{a}_{5}}=5\left( 5+2 \right)\]

\[\Rightarrow {{a}_{5}}=35\]

Therefore, the first five terms of \[{{a}_{n}}=n\left( n+2 \right)\] is \[3,8,15,24\] and \[35\] . 

2. Write the first five terms of the sequences whose \[{{n}^{th}}\] term is \[{{a}_{n}}=\frac{n}{n+1}\] .

Ans:

The given equation is \[{{a}_{n}}=\frac{n}{n+1}\] .

Substitute \[n=1\] in the equation.

\[{{a}_{1}}=\frac{1}{1+1}\]

\[\Rightarrow {{a}_{1}}=\frac{1}{2}\]

Similarly substitute \[n=2,3,4\] and \[5\]in the equation.

\[{{a}_{2}}=\frac{2}{2+1}\]

\[\Rightarrow {{a}_{2}}=\frac{2}{3}\]

\[{{a}_{3}}=\frac{3}{3+1}\]

\[\Rightarrow {{a}_{3}}=\frac{3}{4}\]

\[{{a}_{4}}=\frac{4}{4+1}\]

\[\Rightarrow {{a}_{4}}=\frac{4}{5}\]

\[{{a}_{5}}=\frac{5}{5+1}\]

\[\Rightarrow {{a}_{5}}=\frac{5}{6}\]

Therefore, the first five terms of \[{{a}_{n}}=\frac{n}{n+1}\] is \[\frac{1}{2},\frac{2}{3},\frac{3}{4},\frac{4}{5}\] and \[\frac{5}{6}\] . 

3. Write the first five terms of the sequences whose \[{{n}^{th}}\] term is \[{{a}_{n}}={{2}^{n}}\] .

Ans:

The given equation is \[{{a}_{n}}={{2}^{n}}\] .

Substitute \[n=1\] in the equation.

\[{{a}_{1}}={{2}^{1}}\]

\[\Rightarrow {{a}_{1}}=2\]

Similarly substitute \[n=2,3,4\] and \[5\]in the equation.

\[{{a}_{2}}={{2}^{2}}\]

\[\Rightarrow {{a}_{2}}=4\]

\[{{a}_{3}}={{2}^{3}}\]

\[\Rightarrow {{a}_{3}}=8\]

\[{{a}_{4}}={{2}^{4}}\]

\[\Rightarrow {{a}_{4}}=16\]

\[{{a}_{5}}={{2}^{5}}\]

\[\Rightarrow {{a}_{5}}=32\]

Therefore, the first five terms of \[{{a}_{n}}={{2}^{n}}\] is \[2,4,8,16\] and \[32\] . 

4. Write the first five terms of the sequences whose \[{{n}^{th}}\] term is \[{{a}_{n}}=\frac{2n-3}{6}\] .

Ans:

The given equation is \[{{a}_{n}}=\frac{2n-3}{6}\] .

Substitute \[n=1\] in the equation.

\[{{a}_{1}}=\frac{2\left( 1 \right)-3}{6}\]

\[\Rightarrow {{a}_{1}}=-\frac{1}{6}\]

Similarly substitute \[n=2,3,4\] and \[5\]in the equation.

\[{{a}_{2}}=\frac{2\left( 2 \right)-3}{6}\]

\[\Rightarrow {{a}_{2}}=\frac{1}{6}\]

\[{{a}_{3}}=\frac{2\left( 3 \right)-3}{6}\]

\[\Rightarrow {{a}_{3}}=\frac{3}{6}=\frac{1}{2}\]

\[{{a}_{4}}=\frac{2\left( 4 \right)-3}{6}\]

\[\Rightarrow {{a}_{4}}=\frac{5}{6}\]

\[{{a}_{5}}=\frac{2\left( 5 \right)-3}{6}\]

\[\Rightarrow {{a}_{5}}=\frac{7}{6}\]

Therefore, the first five terms of \[{{a}_{n}}=\frac{2n-3}{6}\] is \[-\frac{1}{6},\frac{1}{6},\frac{1}{2},\frac{5}{6}\] and \[\frac{7}{6}\] . 

5. Write the first five terms of the sequences whose \[{{n}^{th}}\] term is \[{{a}_{n}}={{\left( -1 \right)}^{n-1}}{{5}^{n+1}}\] .

Ans:

The given equation is \[{{a}_{n}}={{\left( -1 \right)}^{n-1}}{{5}^{n+1}}\] .

Substitute \[n=1\] in the equation.

\[{{a}_{1}}={{\left( -1 \right)}^{1-1}}{{5}^{1+1}}\]

\[\Rightarrow {{a}_{1}}={{5}^{2}}=25\]

Similarly substitute \[n=2,3,4\] and \[5\]in the equation.

\[{{a}_{2}}={{\left( -1 \right)}^{2-1}}{{5}^{2+1}}\]

\[\Rightarrow {{a}_{2}}=-{{5}^{3}}=-125\]

\[{{a}_{3}}={{\left( -1 \right)}^{3-1}}{{5}^{3+1}}\]

\[\Rightarrow {{a}_{3}}={{5}^{4}}=625\]

\[{{a}_{4}}={{\left( -1 \right)}^{4-1}}{{5}^{4+1}}\]

\[\Rightarrow {{a}_{4}}=-{{5}^{5}}=-3125\]

\[{{a}_{5}}={{\left( -1 \right)}^{5-1}}{{5}^{5+1}}\]

\[\Rightarrow {{a}_{5}}={{5}^{6}}=15625\]

Therefore, the first five terms of \[{{a}_{n}}={{\left( -1 \right)}^{n-1}}{{5}^{n+1}}\] is \[25,-125,625,-3125\] and \[15625\] . 

6. Write the first five terms of the sequences whose \[{{n}^{th}}\] term is \[{{a}_{n}}=n\frac{{{n}^{2}}+5}{4}\] .

Ans:

The given equation is \[{{a}_{n}}=n\frac{{{n}^{2}}+5}{4}\] .

Substitute \[n=1\] in the equation.

\[{{a}_{1}}=1\cdot \frac{{{1}^{2}}+5}{4}\]

\[\Rightarrow {{a}_{1}}=\frac{6}{4}=\frac{3}{2}\]

Similarly substitute \[n=2,3,4\] and \[5\]in the equation.

\[{{a}_{2}}=2\cdot \frac{{{2}^{2}}+5}{4}\]

\[\Rightarrow {{a}_{2}}=\frac{18}{4}=\frac{9}{2}\]

\[{{a}_{3}}=3\cdot \frac{{{3}^{2}}+5}{4}\]

\[\Rightarrow {{a}_{3}}=\frac{42}{4}=\frac{21}{2}\]

\[{{a}_{4}}=4\cdot \frac{{{4}^{2}}+5}{4}\]

\[\Rightarrow {{a}_{4}}=\frac{84}{4}=21\]

\[{{a}_{5}}=5\cdot \frac{{{5}^{2}}+5}{4}\]

\[\Rightarrow {{a}_{5}}=\frac{150}{4}=\frac{75}{2}\]

Therefore, the first five terms of \[{{a}_{n}}=n\frac{{{n}^{2}}+5}{4}\] is \[\frac{3}{2},\frac{9}{2},\frac{21}{2},21\] and \[\frac{75}{2}\] . 

7. Find the \[{{17}^{th}}\] and \[{{24}^{th}}\] term in the following sequence whose \[{{n}^{th}}\] term is \[{{a}_{n}}=4n-3\] .

Ans:

The given equation is \[{{a}_{n}}=4n-3\] .

Substitute \[n=17\] in the equation.

\[{{a}_{17}}=4\left( 17 \right)-3\]

\[\Rightarrow {{a}_{17}}=65\]

Similarly substitute \[n=24\] in the equation.

\[{{a}_{24}}=4\left( 24 \right)-3\]

\[\Rightarrow {{a}_{24}}=93\]

Therefore, the \[{{17}^{th}}\] and \[{{24}^{th}}\] term of \[{{a}_{n}}=4n-3\] is \[65\] and \[93\] respectively. 

8. Find the \[{{7}^{th}}\] term in the following sequence whose \[{{n}^{th}}\] term is \[{{a}_{n}}=\frac{{{n}^{2}}}{{{2}^{n}}}\] .

Ans:

The given equation is \[{{a}_{n}}=\frac{{{n}^{2}}}{{{2}^{n}}}\] .

Substitute \[n=7\] in the equation.

\[{{a}_{7}}=\frac{{{7}^{2}}}{{{2}^{7}}}\]

\[\Rightarrow {{a}_{7}}=\frac{49}{128}\]

Therefore, the \[{{7}^{th}}\] term of \[{{a}_{n}}=\frac{{{n}^{2}}}{{{2}^{n}}}\] is \[\frac{49}{128}\] . 

9. Find the \[{{9}^{th}}\] term in the following sequence whose \[{{n}^{th}}\] term is \[{{a}_{n}}={{\left( -1 \right)}^{n-1}}{{n}^{3}}\] .

Ans:

The given equation is \[{{a}_{n}}={{\left( -1 \right)}^{n-1}}{{n}^{3}}\] .

Substitute \[n=9\] in the equation.

\[{{a}_{9}}={{\left( -1 \right)}^{9-1}}{{9}^{3}}\]

\[\Rightarrow {{a}_{9}}=729\]

Therefore, the \[{{9}^{th}}\] term of \[{{a}_{n}}={{\left( -1 \right)}^{n-1}}{{n}^{3}}\] is \[729\] . 

10. Find the \[{{20}^{th}}\] term in the following sequence whose \[{{n}^{th}}\] term is \[{{a}_{n}}=\frac{n\left( n-2 \right)}{n+3}\] .

Ans:

The given equation is \[{{a}_{n}}=\frac{n\left( n-2 \right)}{n+3}\] .

Substitute \[n=20\] in the equation.

\[{{a}_{20}}=\frac{20\left( 20-2 \right)}{20+3}\]

\[\Rightarrow {{a}_{20}}=\frac{360}{23}\]

Therefore, the \[{{20}^{th}}\] term of \[{{a}_{n}}=\frac{n\left( n-2 \right)}{n+3}\] is \[\frac{360}{23}\] . 

11. Write the first five terms of the following sequence and obtain the corresponding series: \[{{a}_{1}}=3\], \[{{a}_{n}}=3{{a}_{n-1}}+2\] for all \[n>1\] .

Ans:

The given equation is \[{{a}_{n}}=3{{a}_{n-1}}+2\] where \[{{a}_{1}}=3\] and \[n>1\] .

Substitute \[n=2\] and \[{{a}_{1}}=3\] in the equation.

\[{{a}_{2}}=3{{a}_{2-1}}+2=3\left( 3 \right)+2\]

\[\Rightarrow {{a}_{2}}=11\]

Similarly substitute \[n=3,4\] and \[5\] in the equation.

\[{{a}_{3}}=3{{a}_{3-1}}+2=3\left( 11 \right)+2\]

\[\Rightarrow {{a}_{3}}=35\]

\[{{a}_{4}}=3{{a}_{4-1}}+2=3\left( 35 \right)+2\]

\[\Rightarrow {{a}_{4}}=107\]

\[{{a}_{5}}=3{{a}_{5-1}}+2=3\left( 107 \right)+2\]

\[\Rightarrow {{a}_{5}}=323\]

Therefore, the first five terms of \[{{a}_{n}}=3{{a}_{n-1}}+2\] is \[3,11,35,107\] and \[323\] .

The corresponding series obtained from the sequence is \[3+11+35+107+323+...\]

12. Write the first five terms of the following sequence and obtain the corresponding series: \[{{a}_{1}}=-1\], \[{{a}_{n}}=\frac{{{a}_{n-1}}}{n}\], \[n\ge 2\] .

Ans:

The given equation is \[{{a}_{n}}=\frac{{{a}_{n-1}}}{n}\] where \[{{a}_{1}}=-1\] and \[n\ge 2\] .

Substitute \[n=2\] and \[{{a}_{1}}=-1\] in the equation.

\[{{a}_{2}}=\frac{{{a}_{2-1}}}{2}=\frac{-1}{2}\]

\[\Rightarrow {{a}_{2}}=-\frac{1}{2}\]

Similarly substitute \[n=3,4\] and \[5\] in the equation.

\[{{a}_{3}}=\frac{{{a}_{3-1}}}{3}=\frac{{}^{-1}/{}_{2}}{3}\]

\[\Rightarrow {{a}_{3}}=-\frac{1}{6}\]

\[{{a}_{4}}=\frac{{{a}_{4-1}}}{4}=\frac{{}^{-1}/{}_{6}}{4}\]

\[\Rightarrow {{a}_{4}}=-\frac{1}{24}\]

\[{{a}_{5}}=\frac{{{a}_{5-1}}}{5}=\frac{{}^{-1}/{}_{24}}{5}\]

\[\Rightarrow {{a}_{5}}=-\frac{1}{120}\]

Therefore, the first five terms of \[{{a}_{n}}=\frac{{{a}_{n-1}}}{n}\] is \[-1,-\frac{1}{2},-\frac{1}{6},-\frac{1}{24}\] and \[-\frac{1}{120}\] .

The corresponding series obtained from the sequence is \[\left( -1 \right)+\left( -\frac{1}{2} \right)+\left( -\frac{1}{6} \right)+\left( -\frac{1}{24} \right)+\left( -\frac{1}{120} \right)...\]

13. Write the first five terms of the following sequence and obtain the corresponding series: \[{{a}_{1}}={{a}_{2}}=2\], \[{{a}_{n}}={{a}_{n-1}}-1\], \[n>2\] .

Ans:

The given equation is \[{{a}_{n}}={{a}_{n-1}}-1\] where \[{{a}_{1}}={{a}_{2}}=2\] and \[n>2\] .

Substitute \[n=3\] and \[{{a}_{2}}=2\] in the equation.

\[{{a}_{3}}={{a}_{3-1}}-1=2-1\]

\[\Rightarrow {{a}_{3}}=1\]

Similarly substitute \[n=4\] and \[5\] in the equation.

\[{{a}_{4}}={{a}_{4-1}}-1=1-1\]

\[\Rightarrow {{a}_{4}}=0\]

\[{{a}_{5}}={{a}_{5-1}}-1=0-1\]

\[\Rightarrow {{a}_{5}}=-1\]

Therefore, the first five terms of \[{{a}_{n}}={{a}_{n-1}}-1\] is \[2,2,1,0\] and \[-1\] .

The corresponding series obtained from the sequence is \[2+2+1+0+\left( -1 \right)+...\]

14. The Fibonacci sequence is defined by \[1={{a}_{1}}={{a}_{2}}\], \[{{a}_{n}}={{a}_{n-1}}+{{a}_{n-2}}\] , \[n>2\] . Find \[\frac{{{a}_{n+1}}}{{{a}_{n}}}\], for \[n=1,2,3,4,5\].

Ans:

The given equation is \[{{a}_{n}}={{a}_{n-1}}+{{a}_{n-2}}\] where \[1={{a}_{1}}={{a}_{2}}\] and \[n>2\] .

Substitute \[n=3\] and \[1={{a}_{1}}={{a}_{2}}\] in the equation.

\[{{a}_{3}}={{a}_{3-1}}+{{a}_{3-2}}=1+1\]

\[\Rightarrow {{a}_{3}}=2\]

Similarly substitute \[n=4,5\] and \[6\] in the equation.

\[{{a}_{4}}={{a}_{4-1}}+{{a}_{4-2}}=2+1\]

\[\Rightarrow {{a}_{4}}=3\]

\[{{a}_{5}}={{a}_{5-1}}+{{a}_{5-2}}=3+2\]

\[\Rightarrow {{a}_{5}}=5\]

\[{{a}_{6}}={{a}_{6-1}}+{{a}_{6-2}}=5+3\]

\[\Rightarrow {{a}_{6}}=8\]

Substitute the values of  \[{{a}_{1}}\] and \[{{a}_{2}}\] in the expression \[\frac{{{a}_{n+1}}}{{{a}_{n}}}\] for \[n=1\] .

\[\Rightarrow \frac{{{a}_{1+1}}}{{{a}_{1}}}=\frac{1}{1}=1\]

Similarly, when \[n=2\],

\[\Rightarrow \frac{{{a}_{2+1}}}{{{a}_{2}}}=\frac{2}{1}=2\]

When \[n=3\],

\[\Rightarrow \frac{{{a}_{3+1}}}{{{a}_{3}}}=\frac{3}{2}\]

When \[n=4\],

\[\Rightarrow \frac{{{a}_{4+1}}}{{{a}_{4}}}=\frac{5}{3}\]

When \[n=5\],

\[\Rightarrow \frac{{{a}_{5+1}}}{{{a}_{5}}}=\frac{8}{5}\]

Therefore, the value of \[\frac{{{a}_{n+1}}}{{{a}_{n}}}\] for \[n=1,2,3,4,5\] is \[1,2,\frac{3}{2},\frac{5}{3}\] and \[\frac{8}{5}\] respectively.

Exercise 9.2

1. Find the sum of odd integers from \[1\] to \[2001\] .

Ans:

\[1,3,5,...,1999,2001\] are the odd integers from \[1\] to \[2001\] . An A.P. is formed by this sequence.

The first term of the A.P. is \[a=1\] and the common difference is \[d=2\] .

The \[{{n}^{th}}\] term of the A.P. is given by the equation \[{{a}_{n}}=a+\left( n-1 \right)d\] .

Therefore, \[a+\left( n-1 \right)d=2001\]

Substitute \[a=1\] and \[d=2\] in the equation.

\[\Rightarrow 1+\left( n-1 \right)2=2001\]

\[\Rightarrow 2n-1=2001\]

\[\Rightarrow n=\frac{2001+1}{2}\]

\[\Rightarrow n=1001\]

The sum of first \[n\] terms of an arithmetic progression is given by the equation \[{{S}_{n}}=\frac{n}{2}\left[ 2a+\left( n-1 \right)d \right]\] .

Substitute the values of \[n\], \[a\] and \[d\] in the equation.

\[{{S}_{n}}=\frac{1001}{2}\left[ 2\left( 1 \right)+\left( 1001-1 \right)2 \right]\]

\[=\frac{1001}{2}\left[ 2+\left( 1000 \right)2 \right]\]

\[=\frac{1001}{2}\times 2002\]

\[=1001\times 1001\]

\[=1002001\]

Therefore, \[1002001\] is the sum of the odd integers from \[1\] to \[2001\] .

2. Find the sum of all natural numbers lying between \[100\] and \[1000\], which are multiples of \[5\].

Ans:

\[105,110,...,990,995\] are the natural numbers lying between \[100\] and \[1000\], which are multiples of \[5\]. An A.P. is formed by this sequence.

The first term of the A.P. is \[a=105\] and the common difference is \[d=5\] .

The \[{{n}^{th}}\] term of the A.P. is given by the equation \[{{a}_{n}}=a+\left( n-1 \right)d\] .

Therefore, \[a+\left( n-1 \right)d=995\]

Substitute \[a=105\] and \[d=5\] in the equation.

\[\Rightarrow 105+\left( n-1 \right)5=995\]

\[\Rightarrow 5n+100=995\]

\[\Rightarrow n=\frac{995-100}{5}\]

\[\Rightarrow n=179\]

The sum of first \[n\] terms of an arithmetic progression is given by the equation \[{{S}_{n}}=\frac{n}{2}\left[ 2a+\left( n-1 \right)d \right]\] .

Substitute the values of \[n\], \[a\] and \[d\] in the equation.

\[{{S}_{n}}=\frac{179}{2}\left[ 2\left( 105 \right)+\left( 179-1 \right)5 \right]\]

\[=\frac{179}{2}\left[ 2\left( 105 \right)+\left( 178 \right)5 \right]\]

\[=179\left[ 105+\left( 89 \right)5 \right]\]

\[=179\times 550\]

\[=98450\]

Therefore, \[98450\] is the sum of the natural numbers lying between \[100\] and \[1000\], which are multiples of \[5\].

3. In an A.P., the first term is \[2\] and the sum of the first five terms is one-fourth of the next five terms. Show that \[{{20}^{th}}\] term is \[-112\] . 

Ans:

The first term of the A.P. is \[2\] and let the common difference of the A.P. be \[d\].

Then \[2,2+d,2+2d,2+3d,...\] is the A.P.

\[10+10d\] is the sum of the first five terms and \[10+35d\] is the sum of the next five terms.

According to the conditions given in the question,

\[10+10d=\frac{1}{4}\left( 10+35d \right)\]

\[\Rightarrow 40+40d=10+35d\]

\[\Rightarrow 30=-5d\]

\[\Rightarrow d=-6\]

The \[{{20}^{th}}\] term \[{{a}_{20}}=a+\left( 20-1 \right)d\].

Substitute the values of \[a\] and \[d\] in the equation to obtain \[{{a}_{20}}\] .

\[{{a}_{20}}=2+\left( 20-1 \right)\left( -6 \right)\]

\[=2-114\]

\[=-112\]

Therefore, \[-112\] is the \[{{20}^{th}}\] term of the A.P.

4. How many terms if the A.P. \[-6,-\frac{11}{2},-5,...\] are needed to give the sum \[-25\] ?

Ans:

\[-25\] is the sum of \[n\] terms of the given A.P.

The common difference of the A.P. is \[d=-\frac{11}{2}+6=\frac{-11+12}{2}=\frac{1}{2}\]

The sum of first \[n\] terms of an arithmetic progression is given by the equation \[{{S}_{n}}=\frac{n}{2}\left[ 2a+\left( n-1 \right)d \right]\] .

Substitute \[{{S}_{n}}=-25\], \[a=-6\] and \[d=\frac{1}{2}\] in the equation.

\[-25=\frac{n}{2}\left[ 2\left( -6 \right)+\left( n-1 \right)\frac{1}{2} \right]\]

\[\Rightarrow -50=n\left[ -12+\frac{n}{2}-\frac{1}{2} \right]\]

\[\Rightarrow -50=n\left[ -\frac{25}{2}+\frac{n}{2} \right]\]

\[\Rightarrow -100=n\left[ -25+n \right]\]

\[\Rightarrow {{n}^{2}}-25n+100=0\]

Factorize the equation. 

\[\Rightarrow {{n}^{2}}-5n+20n+100=0\]

\[\Rightarrow n\left( n-5 \right)-20\left( n-5 \right)=0\]

\[\Rightarrow n=20\] or \[5\]

Therefore, \[5\]or \[20\] terms of the A.P. are needed to give the sum \[-25\] .  

5. In an A.P., if \[{{p}^{th}}\] term is \[{1}/{q}\;\] and \[{{q}^{th}}\] term is \[{1}/{p}\;\] , prove that the sum of first \[pq\] terms is \[\frac{1}{2}\left( pq+1 \right)\] , where \[p\ne q\] .

Ans:

The general term of an A.P. is given by the equation \[{{a}_{n}}=a+\left( n-1 \right)d\] .

According to the conditions given in the question,

\[{{p}^{th}}\] term can be written as \[{{a}_{p}}=a+\left( p-1 \right)d=\frac{1}{q}\]  

and \[{{q}^{th}}\] term can be written as \[{{a}_{q}}=a+\left( q-1 \right)d=\frac{1}{p}\]

Subtract \[{{a}_{q}}\] from \[{{a}_{p}}\] .

\[\left( p-1 \right)d-\left( q-1 \right)d=\frac{1}{q}-\frac{1}{p}\]

\[\Rightarrow \left( p-1-q+1 \right)d=\frac{p-q}{pq}\]

\[\Rightarrow \left( p-q \right)d=\frac{p-q}{pq}\]

\[\Rightarrow d=\frac{1}{pq}\]

Substitute \[d=\frac{1}{pq}\] in \[a+\left( p-1 \right)d=\frac{1}{q}\] .

\[a+\left( p-1 \right)\frac{1}{pq}=\frac{1}{q}\]

\[\Rightarrow a=\frac{1}{q}-\frac{1}{q}+\frac{1}{pq}=\frac{1}{pq}\]

The sum of first \[n\] terms of an arithmetic progression is given by the equation \[{{S}_{n}}=\frac{n}{2}\left[ 2a+\left( n-1 \right)d \right]\] .

Substitute the values of \[n\], \[a\] and \[d\] in the equation.

\[{{S}_{pq}}=\frac{pq}{2}\left[ \frac{2}{pq}+\left( pq-1 \right)\frac{1}{pq} \right]\]

\[=1+\frac{1}{2}\left( pq-1 \right)\]

\[=\frac{1}{2}pq+1-\frac{1}{2}\]

\[=\frac{1}{2}pq+\frac{1}{2}\]

\[=\frac{1}{2}\left( pq+1 \right)\]

Therefore, \[\frac{1}{2}\left( pq+1 \right)\] is the sum of first \[pq\] terms of the A.P.

6. If the sum of a certain number of terms of the A.P. \[25,22,19,...\] is \[116\] . Find the last term. 

Ans:

\[116\] is the sum of \[n\] terms of the given A.P.

The common difference of the A.P. is \[d=22-25=-3\]

The sum of first \[n\] terms of an arithmetic progression is given by the equation \[{{S}_{n}}=\frac{n}{2}\left[ 2a+\left( n-1 \right)d \right]\] .

Substitute \[{{S}_{n}}=116\], \[a=25\] and \[d=-3\] in the equation.

\[116=\frac{n}{2}\left[ 2\left( 25 \right)+\left( n-1 \right)-3 \right]\]

\[\Rightarrow 232=n\left[ 50+-3n+3 \right]\]

\[\Rightarrow 232=n\left( 53-3n \right)\]

\[\Rightarrow 3{{n}^{2}}-53n+232=0\]

Factorize the equation. 

\[\Rightarrow 3{{n}^{2}}-24n-29n+232=0\]

\[\Rightarrow 3n\left( n-8 \right)-29\left( n-8 \right)=0\]

\[\Rightarrow \left( n-8 \right)\left( 3n-29 \right)=0\]

\[\Rightarrow n=8\] or \[\frac{29}{3}\]

\[n=8\] as \[n\] cannot be equal to \[\frac{29}{3}\] .

As \[n=8\] the last term is 

\[{{a}_{8}}=a+\left( 8-1 \right)d\]

Substitute \[a=25\] and \[d=-3\] in the equation.

\[{{a}_{8}}=25+\left( 7 \right)\left( -3 \right)\]

\[=25-21\]

\[=4\]

Therefore, \[4\] is the last term of the A.P. 

7. Find the sum to \[n\] terms of the A.P., whose \[{{k}^{th}}\] term is \[5k+1\] .

Ans:

\[5k+1\] is given as the \[{{k}^{th}}\] term of the A.P.

The equation for \[{{k}^{th}}\] term of an A.P. is given as \[{{a}_{k}}+\left( k-1 \right)d\] .

Then,

\[a+\left( k-1 \right)d=5k+1\]

\[\Rightarrow a+kd-d=5k+1\]

By comparing the coefficient of \[k\] we get the value of \[d\] as \[5\]. 

\[\Rightarrow a-d=1\]

\[\Rightarrow a-5=1\]

\[\Rightarrow a=6\]

The sum of first \[n\] terms of an arithmetic progression is given by the equation \[{{S}_{n}}=\frac{n}{2}\left[ 2a+\left( n-1 \right)d \right]\] .

Substitute the values of \[a\] and \[d\] in the equation.

\[{{S}_{n}}=\frac{n}{2}\left[ 2\left( 6 \right)+\left( n-1 \right)5 \right]\]

\[=\frac{n}{2}\left[ 12+5n-5 \right]\]

\[=\frac{n}{2}\left[ 5n+7 \right]\]

Therefore, \[\frac{n}{2}\left[ 5n+7 \right]\] is the sum of \[n\] terms of the A.P.

8. If the sum of \[n\] terms of an A.P. is \[\left( pn+q{{n}^{2}} \right)\], where \[p\] and \[q\] are constants, find the common difference.

Ans:

The sum of first \[n\] terms of an arithmetic progression is given by the equation \[{{S}_{n}}=\frac{n}{2}\left[ 2a+\left( n-1 \right)d \right]\] .

According to the conditions given in the question,

\[\frac{n}{2}\left[ 2a+\left( n-1 \right)d \right]=pn+q{{n}^{2}}\]

\[\Rightarrow \frac{n}{2}\left[ 2a+nd-d \right]=pn+q{{n}^{2}}\]

\[\Rightarrow na+{{n}^{2}}\frac{d}{2}+n\cdot \frac{d}{2}=pn+q{{n}^{2}}\]

By comparing the coefficients of \[{{n}^{2}}\] on both sides, we get the value of \[d\].

That is, \[\frac{d}{2}=q\]

\[\Rightarrow d=2q\]

Therefore, \[2q\] is the common difference of the A.P.

9. The sums of \[n\] terms of two arithmetic progressions are in the ratio \[5n+4:9n+6\]. Find the ratio of their \[{{18}^{th}}\] terms.

Ans:

Let \[{{a}_{1}}\] and \[{{d}_{1}}\] be the first term and the common difference of the first arithmetic progression respectively and \[{{a}_{2}}\] and \[{{d}_{2}}\] be the first term and the common difference of the second arithmetic progression respectively.

According to the conditions given in the question,

Sum of \[n\] terms of first A.P / Sum of \[n\] terms of second A.P. \[=\frac{5n+4}{9n+6}\]

\[\Rightarrow \frac{\frac{n}{2}\left[ 2{{a}_{1}}+\left( n-1 \right){{d}_{1}} \right]}{\frac{n}{2}\left[ 2{{a}_{2}}+\left( n-1 \right){{d}_{2}} \right]}=\frac{5n+4}{9n+6}\]

\[\Rightarrow \frac{2{{a}_{1}}+\left( n-1 \right){{d}_{1}}}{2{{a}_{2}}+\left( n-1 \right){{d}_{2}}}=\frac{5n+4}{9n+6}\]

Substitute \[n=35\] in the equation.

\[\Rightarrow \frac{2{{a}_{1}}+34{{d}_{1}}}{2{{a}_{2}}+34{{d}_{2}}}=\frac{5\left( 35 \right)+4}{9\left( 35 \right)+6}\]

\[\Rightarrow \frac{{{a}_{1}}+17{{d}_{1}}}{{{a}_{2}}+17{{d}_{2}}}=\frac{179}{321}\]

We also know that 

\[{{18}^{th}}\] term of first A.P. / \[{{18}^{th}}\] term of second A.P. \[=\frac{{{a}_{1}}+17{{d}_{1}}}{{{a}_{2}}+17{{d}_{2}}}\]

Then,

\[{{18}^{th}}\] term of first A.P. / \[{{18}^{th}}\] term of second A.P. \[=\frac{179}{321}\]

Therefore, \[179:321\] is the ratio of \[{{18}^{th}}\] term of both the arithmetic progressions.

10. If the sum of first \[p\] terms of an A.P. is equal to the sum of the first \[q\] terms, then find the sum of the first \[\left( p+q \right)\] terms.

Ans:

Let \[a\] be the first term and \[d\] be the common difference of the A.P.

The sum of first \[p\] terms of the A.P. is given by the equation \[{{S}_{p}}=\frac{p}{2}\left[ 2a+\left( p-1 \right)d \right]\] and the first \[q\] terms by \[{{S}_{q}}=\frac{q}{2}\left[ 2a+\left( q-1 \right)d \right]\] .

According to the conditions given in the question,

\[\frac{p}{2}\left[ 2a+\left( p-1 \right)d \right]=\frac{q}{2}\left[ 2a+\left( q-1 \right)d \right]\]

\[\Rightarrow p\left[ 2a+\left( p-1 \right)d \right]=q\left[ 2a+\left( q-1 \right)d \right]\]

\[\Rightarrow 2ap+pd\left( p-1 \right)=2aq+qd\left( q-1 \right)\]

\[\Rightarrow 2ap\left( p-q \right)+d\left[ p\left( p-1 \right)-q\left( q-1 \right)d \right]=0\]

\[\Rightarrow 2ap\left( p-q \right)+d\left[ {{p}^{2}}-p-{{q}^{2}}+q \right]=0\]

\[\Rightarrow 2ap\left( p-q \right)+d\left[ \left( p-q \right)\left( p+q \right)-\left( p-q \right) \right]=0\]

\[\Rightarrow 2ap\left( p-q \right)+d\left[ \left( p-q \right)\left( p+q-1 \right) \right]=0\]

\[\Rightarrow 2a+d\left( p+q-1 \right)=0\]

\[\Rightarrow d=\frac{-2a}{p+q-1}\]

Therefore, sum of first \[p+q\] terms of the A.P. is given by the equation \[{{S}_{p+q}}=\frac{p+q}{2}\left[ 2a+\left( p+q-1 \right)d \right]\] .

Substituting the value of \[d\] in the equation we get,

\[{{S}_{p+q}}=\frac{p+q}{2}\left[ 2a+\left( p+q-1 \right)\left( \frac{-2a}{p+q-1} \right) \right]\]

\[=\frac{p+q}{2}\left[ 2a-2a \right]\]

\[=0\]

Therefore, \[0\] is the sum of the first \[\left( p+q \right)\] terms of the A.P.

11. Sum of the first \[p\], \[q\] and \[r\] terms of an A.P. are \[a\], \[b\] and \[c\], respectively. Prove that \[\frac{a}{p}\left( q-r \right)+\frac{b}{q}\left( r-p \right)+\frac{c}{r}\left( p-q \right)=0\] .

Ans:

Let \[{{a}_{1}}\] be the first term and \[d\] be the common difference of the A.P.

According to the conditions given in the question,

\[{{S}_{p}}=\frac{p}{2}\left[ 2{{a}_{1}}+\left( p-1 \right)d \right]=a\]

\[\Rightarrow 2{{a}_{1}}+\left( p-1 \right)d=\frac{2a}{p}\]

\[{{S}_{q}}=\frac{q}{2}\left[ 2{{a}_{1}}+\left( q-1 \right)d \right]=b\]

\[\Rightarrow 2{{a}_{1}}+\left( q-1 \right)d=\frac{2b}{q}\]

\[{{S}_{r}}=\frac{r}{2}\left[ 2{{a}_{1}}+\left( r-1 \right)d \right]=c\]

\[\Rightarrow 2{{a}_{1}}+\left( r-1 \right)d=\frac{2c}{r}\]

Subtract \[{{S}_{q}}\] from \[{{S}_{p}}\] .

\[\left( p-1 \right)d-\left( q-1 \right)d=\frac{2a}{p}-\frac{2b}{q}\]

\[\Rightarrow d\left( p-1-q+1 \right)=\frac{2aq-2bp}{pq}\]

\[\Rightarrow d\left( p-q \right)=\frac{2aq-2bp}{pq}\]

\[\Rightarrow d=\frac{2\left( aq-bp \right)}{pq\left( p-q \right)}\]

Subtract \[{{S}_{q}}\] from \[{{S}_{p}}\] .

\[\left( q-1 \right)d-\left( r-1 \right)d=\frac{2b}{q}-\frac{2c}{r}\]

\[\Rightarrow d\left( q-1-r+1 \right)=\frac{2b}{q}-\frac{2c}{r}\]

\[\Rightarrow d\left( q-r \right)=\frac{2br-2qc}{qr}\]

\[\Rightarrow d=\frac{2\left( br-qc \right)}{qr\left( q-r \right)}\]

Equate both the values of \[d\].

\[\frac{\left( aq-bp \right)}{pq\left( p-q \right)}=\frac{\left( br-qc \right)}{qr\left( q-r \right)}\]

\[\Rightarrow qr\left( q-r \right)\left( aq-bq \right)=pq\left( p-q \right)\left( br-qc \right)\]

\[\Rightarrow r\left( aq-bq \right)\left( q-r \right)=p\left( br-qc \right)\left( p-q \right)\]

\[\Rightarrow \left( aqr-bqr \right)\left( q-r \right)=\left( brp-qcp \right)\left( p-q \right)\]

Divide both the sides of the equation by \[pqr\].

\[\left( \frac{a}{p}-\frac{b}{q} \right)\left( q-r \right)=\left( \frac{b}{q}-\frac{c}{r} \right)\left( p-q \right)\]

\[\Rightarrow \frac{a}{p}\left( q-r \right)-\frac{b}{q}\left( q-r+p-q \right)+\frac{c}{r}\left( p-q \right)=0\]

\[\Rightarrow \frac{a}{p}\left( q-r \right)-\frac{b}{q}\left( r-p \right)+\frac{c}{r}\left( p-q \right)=0\]

Therefore, \[\frac{a}{p}\left( q-r \right)+\frac{b}{q}\left( r-p \right)+\frac{c}{r}\left( p-q \right)=0\] is proved.

12. The ratio of the sums of \[m\] and \[n\] terms of an A.P. is \[{{m}^{2}}:{{n}^{2}}\]. Show that the ratio of \[{{m}^{th}}\] and \[{{n}^{th}}\] term is \[\left( 2m-1 \right):\left( 2n-1 \right)\] .

Ans:

Let \[a\] be the first term of the A.P. and \[d\] be the common difference.

According to the conditions given in the question,

Sum of \[m\] terms / Sum of \[n\] terms \[=\frac{{{m}^{2}}}{{{n}^{2}}}\]

\[\Rightarrow \frac{\frac{m}{2}\left[ 2a+\left( m-1 \right)d \right]}{\frac{n}{2}\left[ 2a+\left( n-1 \right)d \right]}=\frac{{{m}^{2}}}{{{n}^{2}}}\]

\[\Rightarrow \frac{2a+\left( m-1 \right)d}{2a+\left( n-1 \right)d}=\frac{m}{n}\]

Substitute \[m=2m-1\] and \[n=2n-1\] in the equation.

\[\frac{2a+\left( 2m-2 \right)d}{2a+\left( 2n-2 \right)d}=\frac{2m-1}{2n-1}\]

\[\Rightarrow \frac{a+\left( m-1 \right)d}{a+\left( n-1 \right)d}=\frac{2m-1}{2n-1}\]

We also know that 

\[{{m}^{th}}\] term of A.P. / \[{{n}^{th}}\] term of A.P. \[=\frac{a+\left( m-1 \right)d}{a+\left( n-1 \right)d}\]

Then,

\[{{m}^{th}}\] term of A.P. / \[{{n}^{th}}\] term of A.P. \[=\frac{2m-1}{2n-1}\]

Therefore, the ratio of \[{{m}^{th}}\] and \[{{n}^{th}}\] term is \[\left( 2m-1 \right):\left( 2n-1 \right)\] .

13. If the sum of \[n\] terms of an A.P. is \[3{{n}^{2}}+5n\] and its \[{{m}^{th}}\] term is \[164\], find the value of \[m\].

Ans:

Let \[a\] be the first term of the A.P. and \[d\] be the common difference.

\[{{a}_{m}}=a+\left( m-1 \right)d=164\]

The sum of first \[n\] terms of an arithmetic progression is given by the equation \[{{S}_{n}}=\frac{n}{2}\left[ 2a+\left( n-1 \right)d \right]\] .

According to the conditions given in the question,

\[\frac{n}{2}\left[ 2a+\left( n-1 \right)d \right]=3{{n}^{2}}+5n\]

\[\Rightarrow \frac{n}{2}\left[ 2a+nd-d \right]=3{{n}^{2}}+5n\]

\[\Rightarrow na+{{n}^{2}}\frac{d}{2}+n\cdot \frac{d}{2}=3{{n}^{2}}+5n\]

By comparing the coefficients of \[{{n}^{2}}\] on both sides, we get the value of \[d\].

That is, \[\frac{d}{2}=3\]

\[\Rightarrow d=6\]

By comparing the coefficient of \[n\] on both sides we get the value of \[a\]. 

That is, \[a-\frac{d}{2}=5\]

\[\Rightarrow a-3=5\]

\[\Rightarrow a=8\]

Substitute the values of \[a\] and \[d\] in the equation for \[{{a}_{m}}\] .

\[8+\left( m-1 \right)6=164\]

\[\Rightarrow \left( m-1 \right)6=156\]

\[\Rightarrow \left( m-1 \right)=26\]

\[\Rightarrow m=27\]

Therefore, \[27\] is the value of \[m\].

14. Insert five numbers between \[8\] and \[26\] such that resulting sequence is an A.P.

Ans:

Let the five numbers between \[8\] and \[26\] be \[{{A}_{1}},{{A}_{2}},{{A}_{3}},{{A}_{4}}\] and \[{{A}_{5}}\]. Then the resulting sequence \[8,{{A}_{1}},{{A}_{2}},{{A}_{3}},{{A}_{4}},{{A}_{5}},26\] is an A.P.

The first term of the A.P. \[a=8\], the last term \[b=26\] and the number of terms \[n=7\] .

Substitute the values of \[a,b\] and \[n\] in \[b=a+\left( n-1 \right)d\].

\[\Rightarrow 26=8+\left( 7-1 \right)d\]

\[\Rightarrow 6d=26-8\]

\[\Rightarrow d=3\]

Then substituting the values of \[a\] and \[d\] we get 

\[{{A}_{1}}=a+d=8+3=11\] 

\[{{A}_{2}}=a+2d=8+2\left( 3 \right)=14\]

\[{{A}_{3}}=a+3d=8+3\left( 3 \right)=17\]

\[{{A}_{4}}=a+4d=8+4\left( 3 \right)=20\]

\[{{A}_{5}}=a+5d=8+5\left( 3 \right)=23\]

Therefore, \[11,14,17,20\] and \[23\] are the five numbers between \[8\] and \[26\] .

15. If \[\frac{{{a}^{n}}+{{b}^{n}}}{{{a}^{n-1}}+{{b}^{n-1}}}\] is the A.M. between \[a\] and \[b\] , then find the value of \[n\] .

Ans: 

We know that the A.M. between \[a\] and \[b\] is given by \[\frac{a+b}{2}\] .

According to the conditions given in the question,

\[\frac{a+b}{2}=\frac{{{a}^{n}}+{{b}^{n}}}{{{a}^{n-1}}+{{b}^{n-1}}}\]

\[\Rightarrow \left( a+b \right)\left( {{a}^{n-1}}+{{b}^{n-1}} \right)=2\left( {{a}^{n}}+{{b}^{n}} \right)\]

\[\Rightarrow {{a}^{n}}+a{{b}^{n-1}}+b{{a}^{n-1}}+{{b}^{n}}=2{{a}^{n}}+2{{b}^{n}}\]

\[\Rightarrow a{{b}^{n-1}}+b{{a}^{n-1}}={{a}^{n}}+{{b}^{n}}\]

\[\Rightarrow a{{b}^{n-1}}-{{b}^{n}}={{a}^{n}}-{{a}^{n-1}}b\]

\[\Rightarrow {{b}^{n-1}}\left( a-b \right)={{a}^{n-1}}\left( a-b \right)\]

\[\Rightarrow {{b}^{n-1}}={{a}^{n-1}}\]

\[\Rightarrow {{\left( \frac{a}{b} \right)}^{n-1}}=1={{\left( \frac{a}{b} \right)}^{0}}\]

\[\Rightarrow n-1=0\]

\[\Rightarrow n=1\]

Therefore, the value of \[n\] is \[1\] . 

16. Between \[1\] and \[31\], \[m\] numbers have been inserted in such a way that the  resulting sequence is an A.P. and the ratio of \[{{7}^{th}}\] and \[{{\left( m-1 \right)}^{th}}\]  numbers is \[5:9\] . Find the value of \[m\] .

Ans:

Let the \[m\] numbers between \[1\] and \[31\] be \[{{A}_{1}},{{A}_{2}},...,{{A}_{m}}\] . Then the resulting sequence \[1,{{A}_{1}},{{A}_{2}},...,{{A}_{m}},31\] is an A.P.

The first term of the A.P. \[a=1\], the last term \[b=31\] and the number of terms \[n=m+2\] .

Substitute the values of \[a,b\] and \[n\] in \[b=a+\left( n-1 \right)d\].

\[\Rightarrow 31=1+\left( m+2-1 \right)d\]

\[\Rightarrow 30=\left( m+1 \right)d\]

\[\Rightarrow d=\frac{30}{m+1}\]

We know that

\[{{A}_{1}}=a+d\] 

\[{{A}_{2}}=a+2d\]

\[{{A}_{3}}=a+3d\]

Then, the \[{{7}^{th}}\] and \[{{\left( m-1 \right)}^{th}}\] term is given by the equation,

\[{{A}_{7}}=a+7d\]

\[{{A}_{m-1}}=a+\left( m-1 \right)d\]

According to the conditions given in the question,

\[\frac{a+7d}{a+\left( m-1 \right)d}=\frac{5}{9}\]

\[\Rightarrow \frac{1+7\left( \frac{30}{m+1} \right)}{1+\left( m-1 \right)\left( \frac{30}{m+1} \right)}=\frac{5}{9}\]

\[\Rightarrow \frac{m+1+7\left( 30 \right)}{m+1+30\left( m-1 \right)}=\frac{5}{9}\]

\[\Rightarrow \frac{m+1+210}{m+1+30m-30}=\frac{5}{9}\]

\[\Rightarrow \frac{m+211}{31m-29}=\frac{5}{9}\]

\[\Rightarrow 9m+1899=155m-145\]

\[\Rightarrow 155m-9m=1899+145\]

\[\Rightarrow 146m=2044\]

\[\Rightarrow m=14\]

Therefore, \[14\] is the value of \[m\] .

17. A man starts repaying a loan as first installment of Rs.\[100\] . If he increases the installment by Rs.\[5\] every month, what amount he will pay in the \[{{30}^{th}}\] installment?

Ans:

Rs.\[100\] is the first installment of the load and Rs.\[105\] is the second installment and so on.

Therefore, an A.P. is formed by the amount that the man repays every month.

\[100,105,110...\] is the A.P.

The first term of the A.P. \[a=100\] and the common difference \[d=5\].

The \[{{30}^{th}}\] is given by the equation \[{{A}_{30}}=a+\left( 30-1 \right)d\] .

\[\Rightarrow {{A}_{30}}=100+\left( 29 \right)5\]

\[=100+145\]

\[=245\]

Therefore, Rs.\[245\] is the amount to be paid in the \[{{30}^{th}}\] installment.

18.The difference between any two consecutive interior angles of a polygon is \[5{}^\circ \] . If the smallest angle is \[120{}^\circ \] , find the number of the sides of the polygon.

Ans:

An A.P. is formed by the angles of the polygon. The first term of the A.P. \[a=120{}^\circ \] and the common difference \[d=5{}^\circ \] .

We know that \[180\left( n-2 \right)\] is the sum of all angles of a polygon with \[n\] sides. 

Therefore, \[{{S}_{n}}=180{}^\circ \left( n-2 \right)\]

\[\Rightarrow \frac{n}{2}\left[ 2a+\left( n-1 \right)d \right]=180{}^\circ \left( n-2 \right)\]

\[\Rightarrow \frac{n}{2}\left[ 240{}^\circ +\left( n-1 \right)5{}^\circ  \right]=180{}^\circ \left( n-2 \right)\]

\[\Rightarrow n\left[ 240+\left( n-1 \right)5 \right]=360\left( n-2 \right)\]

\[\Rightarrow 240n+5{{n}^{2}}-5n=360n-720\]

\[\Rightarrow 5{{n}^{2}}-125n+720=0\]

\[\Rightarrow {{n}^{2}}-25n+144=0\]

\[\Rightarrow {{n}^{2}}-16n-9n+144=0\]

\[\Rightarrow n\left( n-16 \right)-9\left( n-16 \right)=0\]

\[\Rightarrow \left( n-9 \right)\left( n-16 \right)=0\]

\[\Rightarrow n=9\] or \[16\]

Therefore, the number of the sides of the polygon \[9\] or \[16\] .

Exercise 9.3

1. Find the \[{{20}^{th}}\] and \[{{n}^{th}}\] term of the G.P. \[\frac{5}{2},\frac{5}{4},\frac{5}{8},...\]

Ans:

\[\frac{5}{2},\frac{5}{4},\frac{5}{8},...\] is the given G.P.

The first term of the G.P. is \[a=\frac{5}{2}\] and the common ratio is \[r=\frac{{5}/{4}\;}{{5}/{2}\;}=\frac{1}{2}\].

The \[{{n}^{th}}\] term of the G.P. is given by the equation \[{{a}_{n}}=a{{r}^{n-1}}\].

Substituting the values of \[a\] and \[r\] we get

\[{{a}_{n}}=\frac{5}{2}{{\left( \frac{1}{2} \right)}^{n-1}}=\frac{5}{\left( 2 \right){{\left( 2 \right)}^{n-1}}}=\frac{5}{{{\left( 2 \right)}^{n}}}\]

Similarly, the \[{{20}^{th}}\] term of the G.P. is \[{{a}_{20}}=a{{r}^{20-1}}\]

\[\Rightarrow {{a}_{20}}=\frac{5}{2}{{\left( \frac{1}{2} \right)}^{19}}=\frac{5}{\left( 2 \right){{\left( 2 \right)}^{19}}}=\frac{5}{{{\left( 2 \right)}^{20}}}\]

Therefore, the \[{{20}^{th}}\] and \[{{n}^{th}}\] term of the given G.P. is \[\frac{5}{{{\left( 2 \right)}^{20}}}\] and \[\frac{5}{{{\left( 2 \right)}^{n}}}\] respectively.

2. Find the \[{{12}^{th}}\] term of a G.P. whose \[{{8}^{th}}\] term is \[192\] and the common ratio is \[2\]. 

Ans:

Let the first term of the G.P. be \[a\] and the common ratio \[r=2\] .

The \[{{8}^{th}}\] term of the G.P. is given by the equation \[{{a}_{8}}=a{{r}^{8-1}}\].

Substituting the values of \[{{a}_{8}}\] and \[r\] we get

\[\Rightarrow 192=a{{\left( 2 \right)}^{7}}\]

\[\Rightarrow {{\left( 2 \right)}^{6}}\left( 3 \right)=a{{\left( 2 \right)}^{7}}\]

\[\Rightarrow a=\frac{{{\left( 2 \right)}^{6}}\left( 3 \right)}{{{\left( 2 \right)}^{7}}}=\frac{3}{2}\]

Then \[{{12}^{th}}\] term of the G.P. is given by the equation \[{{a}_{12}}=a{{r}^{12-1}}\].

Substitute the values of \[a\] and \[r\] in the equation.

\[{{a}_{12}}=\frac{3}{2}{{\left( 2 \right)}^{11}}\]

\[=3{{\left( 2 \right)}^{10}}\]

\[=3072\]

Therefore, the \[{{12}^{th}}\] term of the G.P. is \[3072\] .

3. The \[{{5}^{th}}\], \[{{8}^{th}}\] and \[{{11}^{th}}\] terms of a G.P. are \[p\],\[q\] and \[s\] , respectively. Show that \[{{q}^{2}}=ps\] .  

Ans:

Let the first term and the common ratio of the G.P. be \[a\] and \[r\] respectively.

According to the conditions given in the question,

\[{{a}_{5}}=a{{r}^{5-1}}=a{{r}^{4}}=p\]

\[{{a}_{8}}=a{{r}^{8-1}}=a{{r}^{7}}=q\]

\[{{a}_{11}}=a{{r}^{11-1}}=a{{r}^{10}}=s\]

Dividing \[{{a}_{8}}\] by \[{{a}_{5}}\] we get

\[\frac{a{{r}^{7}}}{a{{r}^{4}}}=\frac{q}{p}\]

\[\Rightarrow {{r}^{3}}=\frac{q}{p}\]

Dividing \[{{a}_{11}}\] by \[{{a}_{8}}\] we get

\[\frac{a{{r}^{10}}}{a{{r}^{7}}}=\frac{s}{q}\]

\[\Rightarrow {{r}^{3}}=\frac{s}{q}\]

Equate both the values of \[{{r}^{3}}\] obtained.

\[\frac{q}{p}=\frac{s}{q}\]

\[\Rightarrow {{q}^{2}}=ps\]

Therefore,  \[{{q}^{2}}=ps\] is proved.

4. The \[{{4}^{th}}\] term of a G.P. is square of its second term, and the first term is \[-3\] . Determine its \[{{7}^{th}}\] term.

Ans:

Let the first term and the common ratio of the G.P. be \[a\] and \[r\] respectively.

It is given that \[a=-3\] .

The \[{{n}^{th}}\] term of the G.P. is given by the equation \[{{a}_{n}}=a{{r}^{n-1}}\].

Then,

\[{{a}_{4}}=a{{r}^{3}}=\left( -3 \right){{r}^{3}}\]

\[{{a}_{2}}=a{{r}^{1}}=\left( -3 \right)r\]

According to the conditions given in the question,

\[\left( -3 \right){{r}^{3}}={{\left[ \left( -3 \right)r \right]}^{2}}\]

\[\Rightarrow -3{{r}^{3}}=9{{r}^{2}}\]

\[\Rightarrow r=-3{{a}_{7}}\]

\[=a{{r}^{6}}\]

\[=\left( -3 \right){{\left( -3 \right)}^{6}}\]

\[=-{{\left( 3 \right)}^{7}}\]

\[=-2187\]

Therefore, \[-2187\] is the seventh term of the G.P.

5. Which term of the following sequences:

1. \[2,2\sqrt{2},4...\] is \[128\] ?

Ans:

\[2,2\sqrt{2},4...\] is the given sequence.

The first term of the G.P. \[a=2\] and the common ratio \[r={\left( 2\sqrt{2} \right)}/{2}\;=\sqrt{2}\] .

\[128\] is the \[{{n}^{th}}\] term of the given sequence.

The \[{{n}^{th}}\] term of the G.P. is given by the equation \[{{a}_{n}}=a{{r}^{n-1}}\].

Therefore, \[a{{r}^{n-1}}=128\]

\[\Rightarrow \left( 2 \right){{\left( \sqrt{2} \right)}^{n-1}}=128\]

\[\Rightarrow \left( 2 \right){{\left( 2 \right)}^{\frac{n-1}{2}}}={{\left( 2 \right)}^{7}}\]

\[\Rightarrow {{\left( 2 \right)}^{\frac{n-1}{2}+1}}={{\left( 2 \right)}^{7}}\]

\[\Rightarrow \frac{n-1}{2}+1=7\]

\[\Rightarrow \frac{n-1}{2}=6\]

\[\Rightarrow n-1=12\]

\[\Rightarrow n=13\]

Therefore, \[128\] is the \[{{13}^{th}}\] term of the given sequence.

2. \[\sqrt{3},3,3\sqrt{3}...\] is \[729\] ?

Ans:

\[\sqrt{3},3,3\sqrt{3}...\] is the given sequence.

The first term of the G.P. \[a=\sqrt{3}\] and the common ratio \[r={3}/{\sqrt{3}}\;=\sqrt{3}\] .

\[729\] is the \[{{n}^{th}}\] term of the given sequence.

The \[{{n}^{th}}\] term of the G.P. is given by the equation \[{{a}_{n}}=a{{r}^{n-1}}\].

Therefore, \[a{{r}^{n-1}}=729\]

\[\Rightarrow \left( \sqrt{3} \right){{\left( \sqrt{3} \right)}^{n-1}}=729\]

\[\Rightarrow {{\left( 3 \right)}^{{1}/{2}\;}}{{\left( 2 \right)}^{\frac{n-1}{2}}}={{\left( 3 \right)}^{6}}\]

\[\Rightarrow {{\left( 3 \right)}^{\frac{1}{2}+\frac{n-1}{2}}}={{\left( 3 \right)}^{6}}\]

\[\Rightarrow \frac{1}{2}+\frac{n-1}{2}=6\]

\[\Rightarrow \frac{1+n-1}{2}=6\]

\[\Rightarrow \frac{n}{2}=6\]

\[\Rightarrow n=12\]

Therefore, \[729\] is the \[{{12}^{th}}\] term of the given sequence.

3. \[\frac{1}{3},\frac{1}{9},\frac{1}{27},...\] is \[\frac{1}{19683}\] ?

Ans:

\[\frac{1}{3},\frac{1}{9},\frac{1}{27},...\] is the given sequence.

The first term of the G.P. \[a=\frac{1}{3}\] and the common ratio \[r=\frac{1}{9}\div \frac{1}{3}=\frac{1}{3}\] .

\[\frac{1}{19683}\] is the \[{{n}^{th}}\] term of the given sequence.

The \[{{n}^{th}}\] term of the G.P. is given by the equation \[{{a}_{n}}=a{{r}^{n-1}}\].

Therefore, \[a{{r}^{n-1}}=\frac{1}{19683}\]

\[\Rightarrow \left( \frac{1}{3} \right){{\left( \frac{1}{3} \right)}^{n-1}}=\frac{1}{19683}\]

\[\Rightarrow {{\left( \frac{1}{3} \right)}^{n}}={{\left( \frac{1}{3} \right)}^{9}}\]

\[\Rightarrow n=9\]

Therefore, \[\frac{1}{19683}\] is the \[{{9}^{th}}\] term of the given sequence.

6. For what values of \[x\] , the numbers \[-\frac{2}{7},x,-\frac{7}{2}\] are in G.P.?

Ans:

\[-\frac{2}{7},x,-\frac{7}{2}\] are the given numbers and the common ratio \[=\frac{x}{-{2}/{7}\;}=\frac{-7x}{2}\]

We also know that, common ratio \[=\frac{-{7}/{2}\;}{x}=\frac{-7}{2x}\]

Equating both the common ratios we get

\[\frac{-7x}{2}=\frac{-7}{2x}\]

\[\Rightarrow {{x}^{2}}=\frac{-2\times 7}{-2\times 7}=1\]

\[\Rightarrow x=\sqrt{1}\]

\[\Rightarrow x=\pm 1\]

Therefore, the given numbers will be in G.P. for \[x=\pm 1\] .

7. Find the sum up to \[20\] terms in the geometric progression \[0.15,0.015,0.0015...\]

Ans:

\[0.15,0.015,0.0015...\] is the given G.P.

The first term of the G.P. \[a=0.15\] and the common ratio \[r=\frac{0.015}{0.15}=0.1\] .

The sum of first \[n\] terms of the G.P. is given by the equation \[{{S}_{n}}=\frac{a\left( 1-{{r}^{n}} \right)}{1-r}\] .

Therefore, the sum of first \[20\] terms of the given G.P. is 

 \[{{S}_{20}}=\frac{0.15\left[ 1-{{\left( 0.1 \right)}^{20}} \right]}{1-0.1}\]

\[=\frac{0.15}{0.9}\left[ 1-{{\left( 0.1 \right)}^{20}} \right]\]

\[=\frac{15}{90}\left[ 1-{{\left( 0.1 \right)}^{20}} \right]\]

\[=\frac{1}{6}\left[ 1-{{\left( 0.1 \right)}^{20}} \right]\]

Therefore, the sum up to \[20\] terms in the geometric progression \[0.15,0.015,0.0015...\] is \[\frac{1}{6}\left[ 1-{{\left( 0.1 \right)}^{20}} \right]\] .

8. Find the sum of \[n\] terms in the geometric progression \[\sqrt{7},\sqrt{21},3\sqrt{7}...\]

Ans:

\[\sqrt{7},\sqrt{21},3\sqrt{7}...\] is the given G.P.

The first term of the G.P. \[a=\sqrt{7}\] and the common ratio \[r=\frac{\sqrt{21}}{\sqrt{7}}=\sqrt{3}\] .

The sum of first \[n\] terms of the G.P. is given by the equation \[{{S}_{n}}=\frac{a\left( 1-{{r}^{n}} \right)}{1-r}\] .

The sum of first \[n\] terms of the given G.P. is 

 \[{{S}_{n}}=\frac{\sqrt{7}\left[ 1-{{\left( \sqrt{3} \right)}^{n}} \right]}{1-\sqrt{3}}\]

\[=\frac{\sqrt{7}\left[ 1-{{\left( \sqrt{3} \right)}^{n}} \right]}{1-\sqrt{3}}\times \frac{1+\sqrt{3}}{1+\sqrt{3}}\]

\[=\frac{\sqrt{7}\left( \sqrt{3}+1 \right)\left[ 1-{{\left( \sqrt{3} \right)}^{n}} \right]}{1-3}\]

\[=\frac{-\sqrt{7}\left( \sqrt{3}+1 \right)\left[ 1-{{\left( \sqrt{3} \right)}^{n}} \right]}{2}\]

\[=\frac{-\sqrt{7}\left( 1+\sqrt{3} \right)}{2}\left[ {{\left( 3 \right)}^{\frac{n}{2}}}-1 \right]\]

Therefore, the sum of \[n\] terms of the geometric progression \[\sqrt{7},\sqrt{21},3\sqrt{7}...\] is \[\frac{-\sqrt{7}\left( 1+\sqrt{3} \right)}{2}\left[ {{\left( 3 \right)}^{\frac{n}{2}}}-1 \right]\] .

9. Find the sum of \[n\] terms in the geometric progression \[1,-a,{{a}^{2}},-{{a}^{3}}...\]( if \[a\ne -1\] )

Ans:

\[1,-a,{{a}^{2}},-{{a}^{3}}...\] is the given G.P.

The first term of the G.P. \[{{a}_{1}}=1\] and the common ratio \[r=-a\] .

The sum of first \[n\] terms of the G.P. is given by the equation \[{{S}_{n}}=\frac{{{a}_{1}}\left( 1-{{r}^{n}} \right)}{1-r}\] .

The sum of first \[n\] terms of the given G.P. is 

 \[{{S}_{n}}=\frac{1\left[ 1-{{\left( -a \right)}^{n}} \right]}{1-\left( -a \right)}\]

\[=\frac{\left[ 1-{{\left( -a \right)}^{n}} \right]}{1+a}\]

Therefore, the sum of \[n\] terms of the geometric progression \[1,-a,{{a}^{2}},-{{a}^{3}}...\] is \[\frac{\left[ 1-{{\left( -a \right)}^{n}} \right]}{1+a}\] .

10. Find the sum of \[n\] terms in the geometric progression \[{{x}^{3}},{{x}^{5}},{{x}^{7}}...\]( if \[a\ne -1\] )

Ans:

\[{{x}^{3}},{{x}^{5}},{{x}^{7}}...\] is the given G.P.

The first term of the G.P. \[a={{x}^{3}}\] and the common ratio \[r={{x}^{2}}\] .

The sum of first \[n\] terms of the G.P. is given by the equation \[{{S}_{n}}=\frac{{{a}_{1}}\left( 1-{{r}^{n}} \right)}{1-r}\] .

The sum of first \[n\] terms of the given G.P. is 

 \[{{S}_{n}}=\frac{{{x}^{3}}\left[ 1-{{\left( {{x}^{2}} \right)}^{n}} \right]}{1-{{x}^{2}}}\]

\[=\frac{{{x}^{3}}\left[ 1-{{x}^{2}}^{n} \right]}{1-{{x}^{2}}}\]

Therefore, the sum of \[n\] terms of the geometric progression \[{{x}^{3}},{{x}^{5}},{{x}^{7}}...\] is \[\frac{{{x}^{3}}\left[ 1-{{x}^{2}}^{n} \right]}{1-{{x}^{2}}}\] .

11. Evaluate \[\sum\limits_{k=1}^{11}{\left( 2+3k \right)}\]

Ans:

\[\sum\limits_{k=1}^{11}{\left( 2+3k \right)=\sum\limits_{k=1}^{11}{(2)}}+\sum\limits_{k=1}^{11}{(3k)=22+\sum\limits_{k=1}^{11}{\left( {{3}^{k}} \right)}}\] …(1)

We know that,

\[\sum\limits_{k=1}^{11}{({{3}^{k}})={{3}^{1}}+{{3}^{2}}+\ldots +{{3}^{11}}}\]

This sequence \[3,{{3}^{2}},{{3}^{3}},\ldots ,{{3}^{11}}\] forms a G.P. Therefore,

\[{{S}_{n}}=\frac{a\left( {{r}^{n}}-1 \right)}{\left( r-1 \right)}\]

Substituting the values to the above equation we get,

\[\Rightarrow {{S}_{n}}=\frac{3\left[ {{\left( 3 \right)}^{11}}-1 \right]}{\left( 3-1 \right)}\]

\[\Rightarrow {{S}_{n}}=\frac{3}{2}\left( {{3}^{11}}-1 \right)\]

Therefore,

\[\Rightarrow \sum\limits_{k=1}^{11}{{{3}^{k}}}=\frac{3}{2}\left( {{3}^{11}}-1 \right)\]

Substitute this value in equation (1). 

\[\sum\limits_{k=1}^{11}{\left( 2+3k \right)=22+\frac{3}{2}\left( {{3}^{11}}-1 \right)}\]

12. The sum of first three terms of a G.P. is \[\frac{39}{10}\] and their product is \[1\]. Find the common ratio and the terms.

Ans:

Let the first three terms of a G.P. be \[\frac{a}{r},a,ar\]. 

Then, its sum is

\[\frac{a}{r}+a+ar=\frac{39}{10}\]       …(1)

And the product is

\[\left( \frac{a}{r} \right)\left( a \right)\left( ar \right)=1\]      …(2)

Solving equation (2) we will get, 

\[{{a}^{3}}=1\]

Considering the real roots,

\[a=1\]

Substitute the value of \[a\] in the equation. 

\[\frac{1}{r}+1+r=\frac{39}{10}\]

\[\Rightarrow 1+r+{{r}^{2}}=\frac{39}{10}r\]

\[\Rightarrow 10+10r+10{{r}^{2}}=39r\]

\[\Rightarrow 10{{r}^{2}}-29r+10=0\]

\[\Rightarrow 10{{r}^{2}}-25r-4r+10=0\]

\[\Rightarrow 5r\left( 2r-5 \right)-2\left( 2r-5 \right)=0\]

\[\Rightarrow \left( 5r-2 \right)\left( 2r-5 \right)=0\]

\[\Rightarrow r=\frac{2}{5}\] or \[\frac{5}{2}\]

Therefore, \[\frac{5}{2},1\] and \[\frac{2}{5}\] are the first three terms of the G.P.

13. How many terms of G.P. \[3,{{3}^{2}},{{3}^{3}}...\] are needed to give the sum 120?

Ans:

Given G.P. \[3,{{3}^{2}},{{3}^{3}},\ldots ,{{3}^{11}}\].

Let there be \[n\] terms to get the sum as \[120\]. 

Then using the formula, we get,

\[{{S}_{n}}=\frac{a\left( {{r}^{n}}-1 \right)}{\left( r-1 \right)}\]       …(1)

Given that,

\[{{S}_{n}}=120\]

\[a=3\]

\[r=3\]

Substituting the given values in equation (1),

\[{{S}_{n}}=120=\frac{3\left( {{3}^{n}}-1 \right)}{3-1}\]

\[\Rightarrow 120=\frac{3\left( {{3}^{n}}-1 \right)}{2}\]

\[\Rightarrow \frac{120\times 2}{3}={{3}^{n}}-1\]

\[\Rightarrow {{3}^{n}}-1=80\]

\[\Rightarrow {{3}^{n}}=81\]

\[\Rightarrow {{3}^{n}}={{3}^{4}}\]

\[\Rightarrow n=4\]

Therefore, for getting the sum as \[120\] the given G.P. should have \[4\] terms.

14. The sum of first three terms of a G.P. is \[16\] and the sum of the next three terms is \[128\]. Determine the first term, the common ratio, and the sum to \[n\] terms of the G.P.

Ans:

Let \[a,ar,a{{r}^{2}},a{{r}^{3}}...\] be the G.P.

According to the conditions given in the question,

\[a+ar+a{{r}^{2}}=16\]         …(1)

\[a{{r}^{3}}+a{{r}^{4}}+a{{r}^{5}}=128\]   …(2)

Equation (1) and (2)  can also be written as,

\[a\left( 1+r+{{r}^{2}} \right)=16\]

\[a{{r}^{3}}\left( 1+r+{{r}^{2}} \right)=128\]

Divide equation (2) by (1) .

\[\frac{\left( 2 \right)}{\left( 1 \right)}\Rightarrow \frac{a{{r}^{3}}\left( 1+r+{{r}^{2}} \right)}{a\left( 1+r+{{r}^{2}} \right)}=\frac{128}{16}\]

\[\Rightarrow {{r}^{3}}=8\]

\[\Rightarrow r=2\]

Substituting the value of \[r\] in equation (1), we get

\[a\left( 1+r+{{r}^{2}} \right)=16\]

\[\Rightarrow a\left( 1+2+4 \right)=16\]

\[\Rightarrow 7a=16\]

\[\Rightarrow a=\frac{16}{7}\]

Sum of \[n\] terms of the G.P. is,

\[{{S}_{n}}=\frac{a\left( {{r}^{n}}-1 \right)}{\left( r-1 \right)}\]\[\]

\[\Rightarrow {{S}_{n}}=\frac{16}{7}\frac{\left( {{2}^{n}}-1 \right)}{2-1}\]

\[\Rightarrow {{S}_{n}}=\frac{16}{7}\left( {{2}^{n}}-1 \right)\]

Therefore, the first term of the G.P. is \[a=\frac{16}{7}\], the common ratio \[r=2\] and the sum of terms \[{{S}_{n}}=\frac{16}{7}\left( {{2}^{n}}-1 \right)\] .

15. Given a G.P. with \[a=729\] and \[{{7}^{th}}\] term \[64\], determine \[{{S}_{7}}\] .

Ans:

Given that\[a=729\] and \[{{a}_{7}}=64\]

Let the common ratio of the G.P be \[r\]. Then,

\[{{a}_{n}}=a{{r}^{n-1}}\]

\[\Rightarrow {{a}_{7}}=a{{r}^{6-1}}\]

\[\Rightarrow 64=729\left( {{r}^{6}} \right)\]

\[\Rightarrow {{r}^{6}}={{\left( \frac{2}{3} \right)}^{6}}\]

\[\Rightarrow r=\frac{2}{3}\]

We know that,

\[{{S}_{n}}=\frac{a\left( 1-{{r}^{n}} \right)}{\left( 1-r \right)}\]

Therefore,

\[{{S}_{7}}=\frac{729\left( 1-{{\left( \frac{2}{3} \right)}^{7}} \right)}{\left( 1-\frac{2}{3} \right)}\]

\[=729\times 3\left( \frac{{{\left( 3 \right)}^{7}}-{{\left( 2 \right)}^{7}}}{{{\left( 3 \right)}^{7}}} \right)\]

\[={{\left( 3 \right)}^{7}}\left( \frac{{{\left( 3 \right)}^{7}}-{{\left( 2 \right)}^{7}}}{{{\left( 3 \right)}^{7}}} \right)\]

\[={{\left( 3 \right)}^{7}}-{{\left( 2 \right)}^{7}}\]

\[=2187-128\]

\[=2059\]

Therefore, the value of \[{{S}_{7}}\] is \[2059\] . 

16. Find a G.P. for which sum of the first two terms is \[-4\] and the fifth term is \[4\] times the third term.

Ans:

Let \[a\] and \[r\] be the first term and common ratio of the G.P. respectively.

According to the conditions given in  the question,

\[{{a}_{5}}=4\times {{a}_{3}}\]

\[\Rightarrow a{{r}^{4}}=4\times a{{r}^{2}}\]

\[\Rightarrow {{r}^{2}}=4\]

\[\Rightarrow r=\pm 2\]

Given that,

\[{{S}_{2}}=-4=\frac{a\left( 1-{{r}^{2}} \right)}{\left( 1-r \right)}\]

Substituting \[r=2\] in the above equation,

\[-4=\frac{a\left[ 1-{{\left( 2 \right)}^{2}} \right]}{1-2}\]

\[\Rightarrow -4=\frac{a\left( 1-4 \right)}{-1}\]

\[\Rightarrow -4=a\left( 3 \right)\]

\[\Rightarrow a=\frac{-4}{3}\]

Now, taking \[r=-2\] , we get,

\[-4=\frac{a\left[ 1-{{\left( -2 \right)}^{2}} \right]}{1-\left( -2 \right)}\]

\[\Rightarrow -4=\frac{a\left( 1-4 \right)}{1+2}\]

\[\Rightarrow -4=\frac{a\left( -3 \right)}{3}\]

\[\Rightarrow a=4\]

Therefore , \[\frac{-4}{3},\frac{-8}{3},\frac{-16}{3},...\] or \[4,-8,-16,-32...\] is the required G.P. 

17. If the \[{{4}^{th}}\],\[{{10}^{th}}\] and \[{{16}^{th}}\]  terms of a G.P. are \[x,y\] and \[z\] , respectively. Prove that \[x,y,z\] are in G.P.

Ans:

Let the first term of the G.P be \[a\] and the common ratio be \[r\].

According to the conditions given in the question,

\[{{a}_{4}}=a{{r}^{3}}=x\]       …(1)

\[{{a}_{10}}=a{{r}^{9}}=y\]      …(2)

\[{{a}_{16}}=a{{r}^{15}}=z\]      …(3)

Then divide equation (2) by (1) .

\[\frac{y}{x}=\frac{a{{r}^{9}}}{a{{r}^{3}}}\]

\[\Rightarrow \frac{y}{x}={{r}^{6}}\]

Now, divide equation (3) by (1).

\[\frac{z}{y}=\frac{a{{r}^{15}}}{a{{r}^{9}}}\]

\[\Rightarrow \frac{z}{y}={{r}^{6}}\]

Therefore, 

\[\frac{y}{x}=\frac{z}{y}\]

Therefore, it is proved that \[x,y,z\] are in G. P.

18. Find the sum to \[n\] terms of the sequence, \[8,88,888,8888...\]

Ans:

\[8,88,888,8888...\] is the given sequence  

The given sequence is not in G.P. In order to make the sequence in G.P., it has to be changed to the form,

\[{{S}_{n}}=8+88+888+8888+...\] to \[n\] terms

\[=\frac{8}{9}\](\[9+99+999+9999+...\]to \[n\] terms)

\[=\frac{8}{9}\](\[\left( 10-1 \right)+\left( {{10}^{2}}-1 \right)+\left( {{10}^{3}}-1 \right)+\left( {{10}^{4}}-1 \right)+\]to \[n\] terms)

\[=\frac{8}{9}\](\[10+{{10}^{2}}+...n\] terms)\[-\]( \[1+1+1+...n\] terms)

\[=\frac{8}{9}\left[ \frac{10\left( {{10}^{n}}-1 \right)}{10-1}-n \right]\]

\[=\frac{8}{9}\left[ \frac{10\left( {{10}^{n}}-1 \right)}{9}-n \right]\]

\[=\frac{80}{81}\left( {{10}^{n}}-1 \right)-\frac{8}{9}n\]

Therefore, the sum of \[n\] terms the given sequence is \[\frac{80}{81}\left( {{10}^{n}}-1 \right)-\frac{8}{9}n\] .

19. Find the sum of the products of the corresponding terms of the sequences \[2,4,8,16,32\] and \[128,32,8,2,{1}/{2}\;\] .

Ans:

\[2\times 128+4\times 32+8\times 8+16\times 2+32\times \frac{1}{2}\] 

\[=64\left[ 4+2+1+\frac{1}{2}+\frac{1}{{{2}^{2}}} \right]\]

is the required sum.

We can see that, \[4,2,1,\frac{1}{2},\frac{1}{{{2}^{2}}}\] is a G.P.

The first term of the G.P. is \[a=4\] and the common ratio is \[r=\frac{1}{2}\] .

We know that,

\[{{S}_{n}}=\frac{a\left( 1-{{r}^{n}} \right)}{\left( 1-r \right)}\]

Therefore,

\[{{S}_{3}}=\frac{4\left[ 1-{{\left( \frac{1}{2} \right)}^{5}} \right]}{1-\frac{1}{2}}\]

\[=\frac{4\left[ 1-\frac{1}{32} \right]}{\frac{1}{2}}\]

\[=8\left( \frac{32-1}{32} \right)\]

\[=\frac{31}{4}\]

Therefore, the required sum \[=64\left( \frac{31}{4} \right)=\left( 16 \right)\left( 31 \right)=496\] .

20. Show that the products of the corresponding terms of the sequences form \[a,ar,a{{r}^{2}},...a{{r}^{n-1}}\] and \[A,AR,A{{R}^{2}},A{{R}^{n-1}}\]  a G.P. and find the common ratio.

Ans:

The sequence \[aA,arAR,a{{r}^{2}}A{{R}^{2}},...a{{r}^{n-1}}A{{R}^{n-1}}\]  forms a G.P. is to be proved.

Second term / First term \[=\frac{arAR}{aA}=rR\]

Third term / Second term \[=\frac{a{{r}^{2}}A{{R}^{2}}}{aA}=rR\]

Therefore, the \[aA,arAR,a{{r}^{2}}A{{R}^{2}},...a{{r}^{n-1}}A{{R}^{n-1}}\] forms a G.P. and the common ratio is \[rR\].

21. Find four numbers forming a geometric progression in which third term is greater than the first term by \[9\], and the second term is greater than the \[{{4}^{th}}\] by \[18\] .

Ans:

Let the first term be \[a\] and the common ratio be \[r\] of the G.P.

\[{{a}_{1}}=a,{{a}_{2}}=ar,{{a}_{3}}=a{{r}^{2}},{{a}_{4}}=a{{r}^{3}}\]

According to the conditions given in the question,

\[{{a}_{3}}={{a}_{1}}+9\]   

\[\Rightarrow a{{r}^{2}}=a+9\]   

\[\Rightarrow a\left( {{r}^{2}}-1 \right)=9\]      ...(1)

 \[{{a}_{2}}={{a}_{4}}+9\]   

\[\Rightarrow ar=a{{r}^{3}}+18\]   

\[\Rightarrow ar\left( 1-{{r}^{2}} \right)=18\]    ...(2)

Divide (2) by (1).

\[\frac{ar\left( 1-{{r}^{2}} \right)}{a\left( {{r}^{2}}-1 \right)}=\frac{18}{9}\]

\[\Rightarrow -r=2\]

\[\Rightarrow r=-2\]

Substitute \[r=-2\] in equation (1).

\[a\left( 4-1 \right)=9\]

\[\Rightarrow a\left( 3 \right)=9\]

\[\Rightarrow a=3\]

Therefore, \[3,3\left( -2 \right),3{{\left( -2 \right)}^{2}}\] and \[3{{\left( -2 \right)}^{3}}\] ,i.e., \[3,-6,12\] and \[-24\] are the first four numbers of the G.P. 

22. If \[{{p}^{th}}\], \[{{q}^{th}}\] and \[{{r}^{th}}\] terms of a G.P. are \[a,b\] and \[c\] , respectively. Prove that \[{{a}^{q-r}}\cdot {{b}^{r-p}}\cdot {{c}^{p-q}}=1\] .

Ans:

Let the first term be \[A\] and the common ration be \[R\] of the G.P.

According to the conditions given in the question,

\[A{{R}^{p-1}}=a\]

\[A{{R}^{q-1}}=b\]

\[A{{R}^{r-1}}=c\]

Then,

\[{{a}^{q-r}}\cdot {{b}^{r-p}}\cdot {{c}^{p-q}}\]

\[={{A}^{q-r}}\times {{R}^{\left( p-1 \right)\left( q-r \right)}}\times {{A}^{r-p}}\times {{R}^{\left( q-1 \right)\left( r-p \right)}}\times {{A}^{q-r}}\times {{R}^{\left( r-1 \right)\left( p-q \right)}}\]

\[={{A}^{q-r+r-p+p-q}}\times {{R}^{\left( pq-pr-q+r \right)+\left( rq-r+p-pq \right)+\left( pr-p-qr+q \right)}}\]

\[={{A}^{0}}\times {{R}^{0}}\]

\[=1\]

Therefore, \[{{a}^{q-r}}\cdot {{b}^{r-p}}\cdot {{c}^{p-q}}=1\] is proved.

23. If the first and \[{{n}^{th}}\] the term of a G.P. are \[a\] and \[b\] , respectively, and if \[P\] is the product of \[n\] terms, prove that \[{{P}^{2}}={{\left( ab \right)}^{n}}\] .

Ans:

\[a\] is the first term and \[b\] is the last term of the G.P.

Therefore, is the G.P. \[a,ar,a{{r}^{2}},a{{r}^{3}}...a{{r}^{n-1}}\] , where the common ratio is \[r\] .

\[b=a{{r}^{n-1}}\]                …(1)

\[P\] is the product of \[n\] terms. Therefore,

\[P=\left( a \right)\left( ar \right)\left( a{{r}^{2}} \right)...\left( a{{r}^{n-1}} \right)\]

\[=\left( a\times a\times ...a \right)\left( r\times {{r}^{2}}\times ...{{r}^{n-1}} \right)\]

\[={{a}^{n}}{{r}^{1+2+...\left( n-1 \right)}}\]          …(2)

We can see that, \[1,2,...\left( n-1 \right)\] is an A.P. Therefore,

\[1+2+...+\left( n-1 \right)\]

\[=\frac{n-1}{2}\left[ 2+\left( n-1-1 \right)\times 1 \right]\]

\[=\frac{n-1}{2}\left[ 2+n-2 \right]\]

\[=\frac{n\left( n-1 \right)}{2}\]

So, equation (2) can be written as \[P={{a}^{n}}{{r}^{\frac{n\left( n-1 \right)}{2}}}\].

Therefore,

\[{{P}^{2}}={{a}^{2n}}{{r}^{n\left( n-1 \right)}}\]

\[={{\left[ {{a}^{2}}{{r}^{\left( n-1 \right)}} \right]}^{n}}\]

\[={{\left[ a\times a{{r}^{n-1}} \right]}^{n}}\]

Substituting (1) in the equation,

\[{{P}^{2}}={{\left( ab \right)}^{n}}\]

Therefore, \[{{P}^{2}}={{\left( ab \right)}^{n}}\] is proved.

24. Show that the ratio of the sum of first \[n\] terms of a G.P. to the sum of terms from \[{{\left( n+1 \right)}^{th}}\] to \[{{\left( 2n \right)}^{th}}\] term is \[\frac{1}{{{r}^{n}}}\] .

Ans:

Let the first term be \[a\] and the common ration be \[r\] of the G.P.

\[\frac{a\left( 1-{{r}^{n}} \right)}{\left( 1-r \right)}\] is the sum of first \[n\] terms.

From \[{{\left( n+1 \right)}^{th}}\] to \[{{\left( 2n \right)}^{th}}\] term there are \[n\] terms.

From \[{{\left( n+1 \right)}^{th}}\] to \[{{\left( 2n \right)}^{th}}\] term the sum of the terms is 

\[{{S}_{n}}=\frac{{{a}_{n+1}}\left( 1-{{r}^{n}} \right)}{1-r}\] 

\[{{a}^{n+1}}=a{{r}^{n+1-1}}=a{{r}^{n}}\]

Therefore, the required ratio is \[=\frac{a\left( 1-{{r}^{n}} \right)}{1-r}\times \frac{1-r}{a{{r}^{n}}\left( 1-{{r}^{n}} \right)}=\frac{1}{{{r}^{n}}}\]

Therefore, \[\frac{1}{{{r}^{n}}}\] is the ratio of the sum of first \[n\] terms of a G.P. to the sum of terms from \[{{\left( n+1 \right)}^{th}}\] to \[{{\left( 2n \right)}^{th}}\]  term .

25. If \[a,b,c\] and \[d\] are in G.P. show that: \[\left( {{a}^{2}}+{{b}^{2}}+{{c}^{2}} \right)\left( {{b}^{2}}+{{c}^{2}}+{{d}^{2}} \right)={{\left( ab+bc+cd \right)}^{2}}\]

Ans:

Let us assume \[a,b,c,d\] are in G.P.

Therefore,

\[bc=ad\]         …(1)

\[{{b}^{2}}=ac\]           …(2)

\[{{c}^{2}}=bd\]           …(3)

To prove :

\[\left( {{a}^{2}}+{{b}^{2}}+{{c}^{2}} \right)\left( {{b}^{2}}+{{c}^{2}}+{{d}^{2}} \right)={{\left( ab+bc+cd \right)}^{2}}\]

\[R.H.S.\]

\[={{\left( ab+bc+cd \right)}^{2}}\]

Substitute (1) in the equation.

\[={{\left( ab+ad+cd \right)}^{2}}\]

\[={{\left( ab+d\left( a+c \right) \right)}^{2}}\]

\[={{a}^{2}}{{b}^{2}}+2abd\left( a+c \right)+{{d}^{2}}{{\left( a+c \right)}^{2}}\]

\[={{a}^{2}}{{b}^{2}}+2{{a}^{2}}bd+2acbd+{{d}^{2}}\left( {{a}^{2}}+2ac+{{c}^{2}} \right)\]

Substitute (1) and (2) in the equation.

\[={{a}^{2}}{{b}^{2}}+2{{a}^{2}}{{c}^{2}}+2{{b}^{2}}{{c}^{2}}+{{d}^{2}}{{a}^{2}}+{{d}^{2}}{{b}^{2}}+{{d}^{2}}{{b}^{2}}+{{d}^{2}}{{c}^{2}}\]

\[={{a}^{2}}{{b}^{2}}+{{a}^{2}}{{c}^{2}}+{{a}^{2}}{{c}^{2}}+{{b}^{2}}{{c}^{2}}+{{b}^{2}}{{c}^{2}}+{{d}^{2}}{{a}^{2}}+{{d}^{2}}{{b}^{2}}+{{d}^{2}}{{b}^{2}}+{{d}^{2}}{{c}^{2}}\]

\[={{a}^{2}}{{b}^{2}}+{{a}^{2}}{{c}^{2}}+{{a}^{2}}{{d}^{2}}+{{b}^{2}}\times {{b}^{2}}+{{b}^{2}}{{c}^{2}}+{{b}^{2}}{{d}^{2}}+{{c}^{2}}{{b}^{2}}+{{c}^{2}}\times {{c}^{2}}+{{c}^{2}}{{d}^{2}}\]

Substitute (2) and (3) in the equation and rearrange the terms.

\[={{a}^{2}}\left( {{b}^{2}}+{{c}^{2}}+{{d}^{2}} \right)+{{b}^{2}}\left( {{b}^{2}}+{{c}^{2}}+{{d}^{2}} \right)+{{c}^{2}}\left( {{b}^{2}}+{{c}^{2}}+{{d}^{2}} \right)\]

\[=\left( {{a}^{2}}+{{b}^{2}}+{{c}^{2}} \right)\left( {{b}^{2}}+{{c}^{2}}+{{d}^{2}} \right)\]

\[=L.H.S.\]

Therefore, \[L.H.S.=R.H.S.\]

Therefore, \[\left( {{a}^{2}}+{{b}^{2}}+{{c}^{2}} \right)\left( {{b}^{2}}+{{c}^{2}}+{{d}^{2}} \right)={{\left( ab+bc+cd \right)}^{2}}\] is proved.

26. Insert two numbers between \[3\] and \[81\] so that the resulting sequence is G.P.

Ans:

Let the two numbers between \[3\] and \[81\] be \[{{G}_{1}}\] and \[{{G}_{2}}\] such that the series, \[3,{{G}_{1}},{{G}_{2}},81\] , forms a G.P.

Let the first term be \[a\] and the common ration be \[r\] of the G.P.

Therefore,

\[81=\left( 3 \right){{\left( r \right)}^{3}}\]

\[\Rightarrow {{r}^{3}}=27\]

Taking the real roots, we get \[r=3\].

When \[r=3\],

\[{{G}_{1}}=ar=\left( 3 \right)\left( 3 \right)=9\]

\[{{G}_{2}}=a{{r}^{2}}=\left( 3 \right){{\left( 3 \right)}^{2}}=27\]

Therefore, \[9\] and \[27\] are the two required numbers. 

27. Find the value of \[n\] so that \[\frac{{{a}^{n+1}}+{{b}^{n+1}}}{{{a}^{n}}+{{b}^{n}}}\] may be the geometric mean between \[a\] and \[b\] .

Ans:

The geometric mean of \[a\] and \[b\] is \[\sqrt{ab}\] .

According to conditions given in the question,

\[\frac{{{a}^{n+1}}+{{b}^{n+1}}}{{{a}^{n}}+{{b}^{n}}}=\sqrt{ab}\]

Square on both the sides.

\[\frac{{{\left( {{a}^{n+1}}+{{b}^{n+1}} \right)}^{2}}}{{{\left( {{a}^{n}}+{{b}^{n}} \right)}^{2}}}=ab\]

\[\Rightarrow {{a}^{2n+2}}+2{{a}^{n+1}}{{b}^{n+1}}+{{b}^{2n+2}}=\left( ab \right)\left( {{a}^{2n}}+2{{a}^{n}}{{b}^{n}}+{{b}^{2n}} \right)\]

\[\Rightarrow {{a}^{2n+2}}+2{{a}^{n+1}}{{b}^{n+1}}+{{b}^{2n+2}}={{a}^{2n+1}}b+2{{a}^{n+1}}{{b}^{n+1}}+a{{b}^{2n+1}}\]

\[\Rightarrow {{a}^{2n+2}}+{{b}^{2n+2}}={{a}^{2n+1}}b+a{{b}^{2n+1}}\]

\[\Rightarrow {{a}^{2n+2}}-{{a}^{2n+1}}b=a{{b}^{2n+1}}-{{b}^{2n+2}}\]

\[\Rightarrow {{a}^{2n+1}}\left( a-b \right)={{b}^{2n+1}}\left( a-b \right)\]

\[\Rightarrow {{\left( \frac{a}{b} \right)}^{2n+1}}=1={{\left( \frac{a}{b} \right)}^{0}}\]

\[\Rightarrow 2n+1=0\]

\[\Rightarrow n=\frac{-1}{2}\]

28. The sum of two numbers is \[6\] times their geometric mean, show that numbers are in the ratio \[\left( 3+2\sqrt{2} \right):\left( 3-2\sqrt{2} \right)\] .

Ans:

Let \[a\] and \[b\] be the two numbers. 

\[\sqrt{ab}\] is the geometric mean.

According to the conditions given in the question,

\[a+b=6\sqrt{ab}\]               …(1)

\[\Rightarrow {{\left( a+b \right)}^{2}}=36ab\] 

Also,

\[{{\left( a-b \right)}^{2}}={{\left( a+b \right)}^{2}}-4ab=36ab-4ab=32ab\]

\[\Rightarrow a-b=\sqrt{32}\sqrt{ab}\]

\[=4\sqrt{2}\sqrt{ab}\]                    …(2)

Add (1) and (2).

\[2a=\left( 6+4\sqrt{2} \right)\sqrt{ab}\]

\[\Rightarrow a=\left( 3+2\sqrt{2} \right)\sqrt{ab}\]

Substitute \[a=\left( 3+2\sqrt{2} \right)\sqrt{ab}\] in equation (1) .

\[b=6\sqrt{ab}-\left( 3+2\sqrt{2} \right)\sqrt{ab}\]
\[\Rightarrow b=\left( 3-2\sqrt{2} \right)\sqrt{ab}\]

Divide \[a\] by \[b\] .

\[\frac{a}{b}=\frac{\left( 3+2\sqrt{2} \right)\sqrt{ab}}{\left( 3-2\sqrt{2} \right)\sqrt{ab}}=\frac{3+2\sqrt{2}}{3-2\sqrt{2}}\]

Therefore, it is proved that the numbers are in the ratio \[\left( 3+2\sqrt{2} \right):\left( 3-2\sqrt{2} \right)\] .

29. If \[A\] and \[B\] be A.M. and G.M., respectively between two positive numbers, prove that the numbers are \[A\pm \sqrt{\left( A+G \right)\left( A-G \right)}\] .

Ans:

Given: The two positive numbers between A.M. and G.M. are \[A\] and \[G\].

Let \[a\] and \[b\] be these two positive numbers.

Therefore, \[AM=A=\frac{a+b}{2}\]      …(1)

\[GM=G=\sqrt{ab}\]                          …(2)

Simplifying (1) and (2) , we get

\[a+b=2A\]         …(3)

\[ab={{G}^{2}}\]              …(4)

Substituting (3) and (4) in the identity,

\[{{\left( a-b \right)}^{2}}={{\left( a+b \right)}^{2}}-4ab\],

We get

\[{{\left( a-b \right)}^{2}}=4{{A}^{2}}-4{{G}^{2}}=4\left( {{A}^{2}}-{{G}^{2}} \right)\]

\[{{\left( a-b \right)}^{2}}=4\left( A+G \right)\left( A-G \right)\]

\[\left( a-b \right)=2\sqrt{\left( A+G \right)\left( A-G \right)}\]     …(5)

Adding (3) and (5) we get ,

\[2a=2A+2\sqrt{\left( A+G \right)\left( A-G \right)}\]

\[\Rightarrow a=A+\sqrt{\left( A+G \right)\left( A-G \right)}\]

Substitute \[a=A+\sqrt{\left( A+G \right)\left( A-G \right)}\] in equation (3).

\[b=2A-A-\sqrt{\left( A+G \right)\left( A-G \right)}\]

\[=A-\sqrt{\left( A+G \right)\left( A-G \right)}\]

Therefore, \[A\pm \sqrt{\left( A+G \right)\left( A-G \right)}\] are the two numbers.

30. The number of bacteria in a certain culture doubles every hour. If there were \[30\] bacteria present in the culture originally, how many bacteria will be present at the end of \[{{2}^{nd}}\] hour, \[{{4}^{th}}\] hour and \[{{n}^{th}}\] hour?

Ans:

The number of bacteria after every hour will form a G.P. as it is given that the number of bacteria doubles every hour. 

Given: \[a=30\] and \[r=2\]

Therefore,

\[{{a}_{3}}=a{{r}^{2}}=\left( 30 \right){{\left( 2 \right)}^{2}}=120\]

That is, \[120\] will be the number of bacteria at the end of \[{{2}^{nd}}\] hour.

\[{{a}_{5}}=a{{r}^{4}}=\left( 30 \right){{\left( 2 \right)}^{4}}=480\]

That is, \[480\] will be the number of bacteria at the end of \[{{4}^{th}}\] hour.

\[{{a}_{n+1}}=a{{r}^{n}}=\left( 30 \right){{2}^{n}}\]

Therefore, \[30{{\left( 2 \right)}^{n}}\] will be the number of bacteria at the end of \[{{n}^{th}}\] hour.

31. What will Rs.\[500\] amounts to in \[10\] years after its deposit in a bank which pays annual interest rate of \[10%\] compounded annually?

Ans:

Rs.\[500\] is the amount deposited in the bank.

The amount \[=\] Rs.\[500\left( 1+\frac{1}{10} \right)=\] Rs.\[500\left( 1.1 \right)\] , at the end of first year.

The amount \[=\] Rs.\[500\left( 1.1 \right)\left( 1.1 \right)\] , at the end of \[{{2}^{nd}}\] year. 

The amount \[=\] Rs.\[500\left( 1.1 \right)\left( 1.1 \right)\left( 1.1 \right)\] , at the end of \[{{3}^{rd}}\] year and so on. 

Therefore, the amount at the end of \[10\] years 

\[=\] Rs.\[500\left( 1.1 \right)\left( 1.1 \right)...\](\[10\]times)

\[=\] Rs.\[500{{\left( 1.1 \right)}^{10}}\]

32. If A.M. and G.M. of roots of a quadratic equation are \[8\] and \[5\] , respectively, then obtain the quadratic equation.

Ans:

Let \[a\] and \[b\] be the root of the quadratic equation.

According to the conditions given in the question,

\[A.M.=\frac{a+b}{2}=8\]

\[\Rightarrow a+b=16\]          …(1)

\[G.M.=\sqrt{ab}=5\]

\[\Rightarrow ab=25\]             …(2)

The quadratic equation is given by the equation,

\[{{x}^{2}}-x\](Sum of roots) \[+\] (Product of roots) \[=0\]

\[{{x}^{2}}-x\left( a+b \right)+\left( ab \right)=0\]

Substituting (1) and (2) in the equation.

\[{{x}^{2}}-16x+25=0\]

Therefore, \[{{x}^{2}}-16x+25=0\] is the required quadratic equation.

Exercise 9.4

1. Find the sum to \[n\] terms of the series \[1\times 2+2\times 3+3\times 4+4\times 5+...\]

Ans:

\[1\times 2+2\times 3+3\times 4+4\times 5+...\] is the given series.

The \[{{n}^{th}}\] term of the series is \[{{a}_{n}}=n\left( n+1 \right)\].

The sum of \[n\] terms of a series is given by the equation \[{{S}_{n}}=\sum\limits_{k=1}^{n}{{{a}_{k}}}\] .

The sum of \[n\] terms of the given series is 

 \[{{S}_{n}}=\sum\limits_{k=1}^{n}{k\left( k+1 \right)}\]

\[=\sum\limits_{k=1}^{n}{{{k}^{2}}}+\sum\limits_{k=1}^{n}{k}\]

\[=\frac{n\left( n+1 \right)\left( 2n+1 \right)}{6}+\frac{n\left( n+1 \right)}{2}\]

\[=\frac{n\left( n+1 \right)}{2}\left( \frac{2n+1}{3}+1 \right)\]

\[=\frac{n\left( n+1 \right)}{2}\left( \frac{2n+4}{3} \right)\]

\[=\frac{n\left( n+1 \right)\left( n+2 \right)}{3}\]

Therefore, the sum of \[n\] terms of the series \[1\times 2+2\times 3+3\times 4+4\times 5+...\] is \[\frac{n\left( n+1 \right)\left( n+2 \right)}{3}\] .

2. Find the sum to \[n\] terms of the series \[1\times 2\times 3+2\times 3\times 4+3\times 4\times 5+...\]

Ans:

\[1\times 2\times 3+2\times 3\times 4+3\times 4\times 5+...\] is the given series.

The \[{{n}^{th}}\] term of the series is 

\[{{a}_{n}}=n\left( n+1 \right)\left( n+2 \right)\]

\[=\left( {{n}^{2}}+n \right)\left( n+2 \right)\]

\[={{n}^{3}}+3{{n}^{2}}+2n\]

The sum of \[n\] terms of a series is given by the equation \[{{S}_{n}}=\sum\limits_{k=1}^{n}{{{a}_{k}}}\] .

The sum of \[n\] terms of the given series is 

 \[{{S}_{n}}=\sum\limits_{k=1}^{n}{\left( {{k}^{3}}+3{{k}^{2}}+2k \right)}\]

\[=\sum\limits_{k=1}^{n}{{{k}^{3}}}+3\sum\limits_{k=1}^{n}{{{k}^{2}}}+2\sum\limits_{k=1}^{n}{k}\]

\[={{\left[ \frac{n\left( n+1 \right)}{2} \right]}^{2}}+\frac{3n\left( n+1 \right)\left( 2n+1 \right)}{6}+\frac{2n\left( n+1 \right)}{2}\]

\[={{\left[ \frac{n\left( n+1 \right)}{2} \right]}^{2}}+\frac{n\left( n+1 \right)\left( 2n+1 \right)}{2}+n\left( n+1 \right)\]

\[=\frac{n\left( n+1 \right)}{2}\left[ \frac{n\left( n+1 \right)}{2}+2n+1+2 \right]\]

\[=\frac{n\left( n+1 \right)}{2}\left[ \frac{{{n}^{2}}+n+4n+6}{2} \right]\]

\[=\frac{n\left( n+1 \right)}{4}\left( {{n}^{2}}+5n+6 \right)\]

\[=\frac{n\left( n+1 \right)}{4}\left( {{n}^{2}}+2n+3n+6 \right)\]

\[=\frac{n\left( n+1 \right)\left[ n\left( n+2 \right)+3\left( n+2 \right) \right]}{4}\]

\[=\frac{n\left( n+1 \right)\left( n+2 \right)\left( n+3 \right)}{4}\]

Therefore, the sum of \[n\] terms of the series \[1\times 2\times 3+2\times 3\times 4+3\times 4\times 5+...\] is \[\frac{n\left( n+1 \right)\left( n+2 \right)\left( n+3 \right)}{4}\] .

3. Find the sum to \[n\] terms of the series \[3\times {{1}^{2}}+5\times {{2}^{2}}+7\times {{3}^{2}}+...\]

Ans:

\[3\times {{1}^{2}}+5\times {{2}^{2}}+7\times {{3}^{2}}+...\] is the given series.

The \[{{n}^{th}}\] term of the series is 

\[{{a}_{n}}=\left( 2n+1 \right){{n}^{2}}\]

\[=2{{n}^{3}}+{{n}^{2}}\]

The sum of \[n\] terms of a series is given by the equation \[{{S}_{n}}=\sum\limits_{k=1}^{n}{{{a}_{k}}}\] .

The sum of \[n\] terms of the given series is 

 \[{{S}_{n}}=\sum\limits_{k=1}^{n}{\left( 2{{k}^{3}}+{{k}^{2}} \right)}\]

\[=2\sum\limits_{k=1}^{n}{{{k}^{3}}}+\sum\limits_{k=1}^{n}{{{k}^{2}}}\]

\[=2{{\left[ \frac{n\left( n+1 \right)}{2} \right]}^{2}}+\frac{n\left( n+1 \right)\left( 2n+1 \right)}{6}\]

\[=\frac{{{n}^{2}}{{\left( n+1 \right)}^{2}}}{2}+\frac{n\left( n+1 \right)\left( 2n+1 \right)}{6}\]

\[=\frac{n\left( n+1 \right)}{2}\left[ n\left( n+1 \right)+\frac{2n+1}{3} \right]\]

\[=\frac{n\left( n+1 \right)}{2}\left[ \frac{3{{n}^{2}}+3n+2n+1}{3} \right]\]

\[=\frac{n\left( n+1 \right)}{2}\left[ \frac{3{{n}^{2}}+5n+1}{3} \right]\]

\[=\frac{n\left( n+1 \right)\left( 3{{n}^{2}}+5n+1 \right)}{6}\]

Therefore, the sum of \[n\] terms of the series \[3\times {{1}^{2}}+5\times {{2}^{2}}+7\times {{3}^{2}}+...\] is \[\frac{n\left( n+1 \right)\left( 3{{n}^{2}}+5n+1 \right)}{6}\] .

4. Find the sum to \[n\] terms of the series \[\frac{1}{1\times 2}+\frac{1}{2\times 3}+\frac{1}{3\times 4}+...\]

Ans:

\[\frac{1}{1\times 2}+\frac{1}{2\times 3}+\frac{1}{3\times 4}+...\] is the given series.

The \[{{n}^{th}}\] term of the series is 

\[{{a}_{n}}=\frac{1}{n\left( n+1 \right)}\]

By partial fractions the above equation can be written as

\[{{a}_{n}}=\frac{1}{n}-\frac{1}{n+1}\]

Therefore,

\[{{a}_{1}}=\frac{1}{1}-\frac{1}{2}\]

\[{{a}_{2}}=\frac{1}{2}-\frac{1}{3}\]

\[{{a}_{3}}=\frac{1}{3}-\frac{1}{4}...\]

\[{{a}_{n}}=\frac{1}{n}-\frac{1}{n+1}\]

Add the above terms.

\[{{a}_{1}}+{{a}_{2}}+...+{{a}_{n}}=\left[ \frac{1}{1}+\frac{1}{2}+\frac{1}{3}+...\frac{1}{n} \right]-\left[ \frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...\frac{1}{n+1} \right]\] 

The sum of \[n\] terms of a series is given by the equation \[{{S}_{n}}=\sum\limits_{k=1}^{n}{{{a}_{k}}}\] .

The sum of \[n\] terms of the given series is 

 \[{{S}_{n}}=1-\frac{1}{n+1}\]

\[=\frac{n+1-1}{n+1}\]

\[=\frac{n}{n+1}\]

Therefore, the sum of \[n\] terms of the series \[\frac{1}{1\times 2}+\frac{1}{2\times 3}+\frac{1}{3\times 4}+...\] is \[\frac{n}{n+1}\] .

5. Find the sum to \[n\] terms of the series \[{{5}^{2}}+{{6}^{2}}+{{7}^{2}}+...+{{20}^{2}}\]

Ans:

\[{{5}^{2}}+{{6}^{2}}+{{7}^{2}}+...+{{20}^{2}}\] is the given series.

The \[{{n}^{th}}\] term of the series is 

\[{{a}_{n}}={{\left( n+4 \right)}^{2}}\]

\[={{n}^{2}}+8n+16\]

The sum of \[n\] terms of a series is given by the equation \[{{S}_{n}}=\sum\limits_{k=1}^{n}{{{a}_{k}}}\] .

The sum of \[n\] terms of the given series is 

 \[{{S}_{n}}=\sum\limits_{k=1}^{n}{\left( {{k}^{2}}+8k+16 \right)}\]

\[=\sum\limits_{k=1}^{n}{{{k}^{2}}}+8\sum\limits_{k=1}^{n}{k}+16\]

\[=\frac{n\left( n+1 \right)\left( 2n+1 \right)}{6}+\frac{8n\left( n+1 \right)}{2}+16n\]

Then,

\[{{20}^{2}}={{\left( 16+4 \right)}^{2}}\] is the \[{{16}^{th}}\] term.

Therefore,

\[{{S}_{16}}=\frac{16\left( 16+1 \right)\left( 2\times 16+1 \right)}{6}+\frac{8\times 16\left( 16+1 \right)}{2}+16\times 16\]

\[=\frac{16\left( 17 \right)\left( 33 \right)}{6}+\frac{\left( 8 \right)\left( 16 \right)\left( 17 \right)}{2}+256\]

\[=1496+1088+256\]

\[=2840\]

Therefore, the sum of \[n\] terms of the series \[{{5}^{2}}+{{6}^{2}}+{{7}^{2}}+...+{{20}^{2}}\] is \[2840\] .

6. Find the sum to \[n\] terms of the series \[3\times 8+6\times 11+9\times 14+...\]

Ans:

\[3\times 8+6\times 11+9\times 14+...{{a}_{n}}\] is the given series.

The \[{{n}^{th}}\] term of the series is 

\[{{a}_{n}}=\] (\[{{n}^{th}}\]term of \[3,6,9...\])\[\times \](\[{{n}^{th}}\]term of \[8,11,14...\])

\[=\left( 3n \right)\left( 3n+5 \right)\]

\[=9{{n}^{2}}+15n\]

The sum of \[n\] terms of a series is given by the equation \[{{S}_{n}}=\sum\limits_{k=1}^{n}{{{a}_{k}}}\] .

The sum of \[n\] terms of the given series is 

 \[{{S}_{n}}=\sum\limits_{k=1}^{n}{\left( 9{{k}^{2}}+15k \right)}\]

\[=9\sum\limits_{k=1}^{n}{{{k}^{2}}}+15\sum\limits_{k=1}^{n}{k}\]

\[=9\times \frac{n\left( n+1 \right)\left( 2n+1 \right)}{6}+15\times \frac{n\left( n+1 \right)}{2}\]

\[=\frac{3n\left( n+1 \right)\left( 2n+1 \right)}{2}+\frac{15n\left( n+1 \right)}{2}\]

\[=\frac{3n\left( n+1 \right)}{2}\left( 2n+1+5 \right)\]

\[=\frac{3n\left( n+1 \right)}{2}\left( 2n+6 \right)\]

\[=3n\left( n+1 \right)\left( n+3 \right)\]

Therefore, the sum of \[n\] terms of the series \[3\times 8+6\times 11+9\times 14+...\] is \[3n\left( n+1 \right)\left( n+3 \right)\] .

7. Find the sum to \[n\] terms of the series \[{{1}^{2}}+\left( {{1}^{2}}\times {{2}^{2}} \right)+\left( {{1}^{2}}\times {{2}^{2}}\times {{3}^{2}} \right)+...\]

Ans:

\[{{1}^{2}}+\left( {{1}^{2}}\times {{2}^{2}} \right)+\left( {{1}^{2}}\times {{2}^{2}}\times {{3}^{2}} \right)+...{{a}_{n}}\] is the given series.

The \[{{n}^{th}}\] term of the series is 

\[{{a}_{n}}=\left( {{1}^{2}}+{{2}^{2}}+{{3}^{2}}+...+{{n}^{2}} \right)\]

\[=\frac{n\left( n+1 \right)\left( n+1 \right)}{6}\]

\[=\frac{n\left( 2{{n}^{2}}+3n+1 \right)}{6}\]

\[=\frac{2{{n}^{3}}+3{{n}^{2}}+n}{6}\]

\[=\frac{1}{3}{{n}^{3}}+\frac{1}{2}{{n}^{2}}+\frac{1}{6}n\]

The sum of \[n\] terms of a series is given by the equation \[{{S}_{n}}=\sum\limits_{k=1}^{n}{{{a}_{k}}}\] .

The sum of \[n\] terms of the given series is 

 \[{{S}_{n}}=\sum\limits_{k=1}^{n}{\left( \frac{1}{3}{{k}^{3}}+\frac{1}{2}{{k}^{2}}+\frac{1}{6}k \right)}\]

\[=\frac{1}{3}\sum\limits_{k=1}^{n}{{{k}^{3}}}+\frac{1}{2}\sum\limits_{k=1}^{n}{{{k}^{2}}}+\frac{1}{6}\sum\limits_{k=1}^{n}{k}\]

\[=\frac{1}{3}\frac{{{n}^{2}}{{\left( n+1 \right)}^{2}}}{{{\left( 2 \right)}^{2}}}+\frac{1}{2}\times \frac{n\left( n+1 \right)\left( 2n+1 \right)}{6}+\frac{1}{6}\times \frac{n\left( n+1 \right)}{2}\]

\[=\frac{n\left( n+1 \right)}{6}\left[ \frac{n\left( n+1 \right)}{2}+\frac{\left( 2n+1 \right)}{2}+\frac{1}{2} \right]\]

\[=\frac{n\left( n+1 \right)}{6}\left[ \frac{{{n}^{2}}+n+2n+1+1}{2} \right]\]

\[=\frac{n\left( n+1 \right)}{6}\left[ \frac{{{n}^{2}}+n+2n+2}{2} \right]\]

\[=\frac{n\left( n+1 \right)}{6}\left[ \frac{\left( n+1 \right)+2\left( n+1 \right)}{2} \right]\]

\[=\frac{n\left( n+1 \right)}{6}\left[ \frac{\left( n+1 \right)\left( n+2 \right)}{2} \right]\]

\[=\frac{n{{\left( n+1 \right)}^{2}}\left( n+2 \right)}{12}\]

Therefore, the sum of \[n\] terms of the series \[{{1}^{2}}+\left( {{1}^{2}}\times {{2}^{2}} \right)+\left( {{1}^{2}}\times {{2}^{2}}\times {{3}^{2}} \right)+...\] is \[\frac{n{{\left( n+1 \right)}^{2}}\left( n+2 \right)}{12}\] .

8. Find the sum to n terms of the series whose  \[{{n}^{th}}\]term is given by \[n\left( n+1 \right)\left( n+4 \right)\]\[\]

Ans:

The \[{{n}^{th}}\] term of the series is 

\[{{a}_{n}}=n\left( n+1 \right)\left( n+4 \right)\]

\[=n\left( {{n}^{2}}+5n+4 \right)\]

\[={{n}^{3}}+5{{n}^{2}}+4n\]

The sum of \[n\] terms of a series is given by the equation \[{{S}_{n}}=\sum\limits_{k=1}^{n}{{{a}_{k}}}\] .

The sum of \[n\] terms of the given series is 

 \[{{S}_{n}}=\sum\limits_{k=1}^{n}{{{k}^{3}}}+5\sum\limits_{k=1}^{n}{{{k}^{2}}+4\sum\limits_{k=1}^{n}{k}}\]

\[=\frac{{{n}^{2}}{{\left( n+1 \right)}^{2}}}{4}+\frac{5n\left( n+1 \right)\left( 2n+1 \right)}{6}+\frac{4n\left( n+1 \right)}{2}\]

\[=\frac{n\left( n+1 \right)}{2}\left[ \frac{n\left( n+1 \right)}{2}+\frac{5\left( 2n+1 \right)}{3}+4 \right]\]

\[=\frac{n\left( n+1 \right)}{2}\left[ \frac{3{{n}^{2}}+3n+20n+10+24}{6} \right]\]

\[=\frac{n\left( n+1 \right)}{2}\left[ \frac{3{{n}^{2}}+23n+34}{6} \right]\]

\[=\frac{n\left( n+1 \right)\left( 3{{n}^{2}}+23n+34 \right)}{12}\]

Therefore, the sum of \[{{n}^{th}}\] terms of the series whose \[{{n}^{th}}\] term is given by  \[n\left( n+1 \right)\left( n+4 \right)\] is \[\frac{n\left( n+1 \right)\left( 3{{n}^{2}}+23n+34 \right)}{12}\] .

9. Find the sum to n terms of the series whose  \[{{n}^{th}}\]term is given by \[{{n}^{2}}+{{2}^{n}}\]\[\]

Ans:

The \[{{n}^{th}}\] term of the series is 

\[{{a}_{n}}={{n}^{2}}+{{2}^{n}}\]

The sum of \[n\] terms of a series is given by the equation \[{{S}_{n}}=\sum\limits_{k=1}^{n}{{{a}_{k}}}\] .

The sum of first \[n\] terms of the given series is 

 \[{{S}_{n}}=\sum\limits_{k=1}^{n}{{{k}^{2}}+{{2}^{k}}}\]

\[=\sum\limits_{k=1}^{n}{{{k}^{2}}}+\sum\limits_{k=1}^{n}{{{2}^{k}}}\]

Let \[\sum\limits_{k=1}^{n}{{{2}^{k}}}={{2}^{1}}+{{2}^{2}}+{{2}^{3}}+...\]

Both the first term and the common ratio of \[{{2}^{1}}+{{2}^{2}}+{{2}^{3}}+...\]which forms a G.P. is \[2\].

Therefore,

\[\sum\limits_{k=1}^{n}{{{2}^{k}}}=\frac{\left( 2 \right)\left[ {{\left( 2 \right)}^{n}}-1 \right]}{2-1}\]

\[=2\left( {{2}^{n}}-1 \right)\]

Then,

\[{{S}_{n}}=\sum\limits_{k=1}^{n}{{{k}^{2}}}+2\left( {{2}^{n}}+1 \right)\]

\[=\frac{n\left( n+1 \right)\left( 2n+1 \right)}{6}+2\left( {{2}^{n}}+1 \right)\]

Therefore, the sum of \[n\] terms of the series whose \[{{n}^{th}}\] term is given by \[{{n}^{2}}+{{2}^{n}}\] is \[\frac{n\left( n+1 \right)\left( 2n+1 \right)}{6}+2\left( {{2}^{n}}+1 \right)\] .

10. Find the sum to n terms of the series whose \[{{n}^{th}}\]term is given by \[{{\left( 2n-1 \right)}^{2}}\]\[\]

Ans:

The \[{{n}^{th}}\] term of the series is 

\[{{a}_{n}}={{\left( 2n-1 \right)}^{2}}\]

\[=4{{n}^{2}}-4n+1\]

The sum of \[n\] terms of a series is given by the equation \[{{S}_{n}}=\sum\limits_{k=1}^{n}{{{a}_{k}}}\] .

The sum of first \[n\] terms of the given series is 

 \[{{S}_{n}}=\sum\limits_{k=1}^{n}{\left( 4{{k}^{2}}-4k+1 \right)}\]

\[=4\sum\limits_{k=1}^{n}{{{k}^{2}}}-4\sum\limits_{k=1}^{n}{k}+\sum\limits_{k=1}^{n}{1}\]

\[=\frac{4n\left( n+1 \right)\left( 2n+1 \right)}{6}-\frac{4n\left( n+1 \right)}{2}+n\]

\[=\frac{2n\left( n+1 \right)\left( 2n+1 \right)}{3}-2n\left( n+1 \right)+n\]

\[=n\left[ \frac{2\left( 2{{n}^{2}}+3n+1 \right)}{3}-2\left( n+1 \right)+1 \right]\]

\[=n\left[ \frac{4{{n}^{2}}+6n+2-6n-6+3}{3} \right]\]

\[=n\left[ \frac{4{{n}^{2}}-1}{3} \right]\]

\[=\frac{n\left( 2n+1 \right)\left( 2n-1 \right)}{3}\]

Therefore, the sum of \[n\] terms of the series whose \[{{n}^{th}}\] term is given by \[{{\left( 2n-1 \right)}^{2}}\] is \[\frac{n\left( 2n+1 \right)\left( 2n-1 \right)}{3}\] .

Miscellaneous Exercise

1. Show that the sum of \[{{\left( m+n \right)}^{th}}\] and \[{{\left( m-n \right)}^{th}}\] terms of an A.P. is equal to twice the \[{{m}^{th}}\] term.

Ans:

Let the first term of the A.P. be \[a\] and the common difference be \[d\] . 

The term of an A.P. is given by the equation

\[{{a}_{k}}=a+\left( k-1 \right)d\] 

Therefore,

\[{{a}_{m+n}}=a+\left( m+n-1 \right)d\]

\[{{a}_{m-n}}=a+\left( m-n-1 \right)d\]

\[{{a}_{m}}=a+\left( m-1 \right)d\]

Add \[{{a}_{m+n}}\] and \[{{a}_{m-n}}\].

\[{{a}_{m+n}}+{{a}_{m-n}}=a+\left( m+n-1 \right)d+a+\left( m-n-1 \right)d\]

\[=2a+\left( m+n-1+m-n-1 \right)d\]

\[=2a+\left( 2m-2 \right)d\]

\[=2a+2\left( m-1 \right)d\]

\[=2\left[ a+\left( m-1 \right)d \right]\]

\[=2{{a}_{m}}\]

Therefore, the sum of \[{{\left( m+n \right)}^{th}}\] and \[{{\left( m-n \right)}^{th}}\] terms of an A.P. is equal to twice the \[{{m}^{th}}\] term is proved.

2. Let the sum of three numbers in A.P., is \[24\] and their product is \[440\], find the numbers. 

Ans:

Let \[a-d\], \[a\] and \[a+d\] be the three numbers in A.P.

According to the conditions given in the question,

\[\left( a-d \right)+\left( a \right)+\left( a+d \right)=24\]

\[\Rightarrow 3a=24\]

\[\Rightarrow a=8\]

\[\left( a-d \right)\left( a \right)\left( a+d \right)=440\]

Substituting \[a=8\] in the equation.

\[\left( 8-d \right)\left( 8 \right)\left( 8+d \right)=440\]

\[\Rightarrow \left( 8-d \right)\left( 8+d \right)=55\]

\[\Rightarrow 64-{{d}^{2}}=55\]

\[\Rightarrow {{d}^{2}}=64-55\]

\[\Rightarrow {{d}^{2}}=9\]

\[\Rightarrow d=\pm 3\]

When \[d=3\],

the three numbers are \[5,8\] and \[11\].

When \[d=-3\],

the three numbers are \[11,8\] and \[5\].

Therefore, \[5,8\] and \[11\] are the three numbers.

3. Let the sum of \[n,2n,3n\]terms of an A.P. be \[{{S}_{1}},{{S}_{2}}\] and \[{{S}_{3}}\], respectively, show that \[{{S}_{3}}=3\left( {{S}_{2}}-{{S}_{1}} \right)\] .

Ans:

Let the first term of the A.P. be \[a\] and the common difference be \[d\] . 

The sum of first \[n\] terms of an arithmetic progression is given by the equation \[{{S}_{n}}=\frac{n}{2}\left[ 2a+\left( n-1 \right)d \right]\] .

Therefore,

\[{{S}_{1}}=\frac{n}{2}\left[ 2a+\left( n-1 \right)d \right]\]

\[{{S}_{2}}=\frac{2n}{2}\left[ 2a+\left( n-1 \right)d \right]=n\left[ 2a+\left( n-1 \right)d \right]\]

\[{{S}_{n}}=\frac{3n}{2}\left[ 2a+\left( n-1 \right)d \right]\]

Subtract \[{{S}_{1}}\] from \[{{S}_{2}}\].

\[{{S}_{2}}-{{S}_{1}}={{S}_{2}}=n\left[ 2a+\left( n-1 \right)d \right]-\frac{n}{2}\left[ 2a+\left( n-1 \right)d \right]\]

\[=n\left[ \frac{4a+4nd-2d-2a-nd+d}{2} \right]\]

\[=n\left[ \frac{2a+3nd-d}{2} \right]\]

\[=\frac{n}{2}\left[ 2a+\left( 3n-1 \right)d \right]\]

Therefore,

\[{{S}_{3}}=\frac{3n}{2}\left[ 2a+\left( 3n-1 \right)d \right]\]

\[=3\left( {{S}_{2}}-{{S}_{1}} \right)\]

Therefore, \[{{S}_{3}}=3\left( {{S}_{2}}-{{S}_{1}} \right)\] is proved.

4. Find the sum of all numbers between \[200\] and \[400\] which are divisible by \[7\].

Ans:

\[203,210,217,...,399\] are the numbers lying between \[200\] and \[400\], which are divisible by \[7\]. An A.P. is formed by this series.

The first term of the A.P. is \[a=203\], the last term is \[{{a}_{n}}=399\] and the common difference is \[d=7\] .

The \[{{n}^{th}}\] term of the A.P. is given by the equation \[{{a}_{n}}=a+\left( n-1 \right)d\] .

Therefore, \[a+\left( n-1 \right)d=399\]

Substitute \[a=203\] and \[d=7\] in the equation.

\[\Rightarrow 203+\left( n-1 \right)7=399\]

\[\Rightarrow \left( n-1 \right)7=196\]

\[\Rightarrow n-1=28\]

\[\Rightarrow n=29\]

The sum of first \[n\] terms of an arithmetic progression is given by the equation \[{{S}_{n}}=\frac{n}{2}\left[ a+{{a}_{n}} \right]\] .

Substitute the values of \[n\], \[a\] and \[d\] in the equation.

\[{{S}_{29}}=\frac{29}{2}\left( 203+399 \right)\]

\[=\frac{29}{2}\left( 602 \right)\]

\[=\left( 29 \right)\left( 301 \right)\]

\[=8729\]

Therefore, \[8729\] is the sum of numbers lying between \[200\] and \[400\], which are divisible by \[7\].

5. Find the sum of integers from \[1\] to \[100\] that are divisible by \[2\] or \[5\].

Ans:

\[2,4,6,...100\] are the integers lying between \[1\] to \[100\], which are divisible by \[2\]. An A.P. is formed by this series.

The first term of the A.P. and the common difference is equal to \[2\] .

The \[{{n}^{th}}\] term of the A.P. is given by the equation \[{{a}_{n}}=a+\left( n-1 \right)d\] .

Substitute \[a=2\] and \[d=2\] in the equation.

\[\Rightarrow 100=2+\left( n-1 \right)2\]

\[\Rightarrow n=50\]

The sum of first \[n\] terms of an arithmetic progression is given by the equation \[{{S}_{n}}=\frac{n}{2}\left[ 2a+\left( n-1 \right)d \right]\] .

Substitute the values of \[n\], \[a\] and \[d\] in the equation.

\[2+4+6+...+100=\frac{50}{2}\left[ 2\left( 2 \right)+\left( 50-1 \right)\left( 2 \right) \right]\]

\[=\frac{50}{2}\left( 4+98 \right)\]

\[=\left( 25 \right)\left( 102 \right)\]

\[=2550\]

\[5,10,...100\] are the integers lying between \[1\] to \[100\], which are divisible by \[5\]. An A.P. is formed by this series.

The first term of the A.P. and the common difference is equal to \[5\] .

The \[{{n}^{th}}\] term of the A.P. is given by the equation \[{{a}_{n}}=a+\left( n-1 \right)d\] .

Substitute \[a=5\] and \[d=5\] in the equation.

\[\Rightarrow 100=5+\left( n-1 \right)5\]

\[\Rightarrow 5n=100\]

\[\Rightarrow n=20\]

The sum of first \[n\] terms of an arithmetic progression is given by the equation \[{{S}_{n}}=\frac{n}{2}\left[ 2a+\left( n-1 \right)d \right]\] .

Substitute the values of \[n\], \[a\] and \[d\] in the equation.

\[5+10+...+100=\frac{20}{2}\left[ 2\left( 5 \right)+\left( 20-1 \right)5 \right]\]

\[=10\left[ 10+\left( 19 \right)5 \right]\]

\[=10\left[ 10+95 \right]\]

\[=10\times 105\]

\[=1050\]

\[10,20,...100\] are the integers lying between \[1\] to \[100\], which are divisible by \[2\] and \[5\]. An A.P. is formed by this series.

The first term of the A.P. and the common difference is equal to \[10\] .

The \[{{n}^{th}}\] term of the A.P. is given by the equation \[{{a}_{n}}=a+\left( n-1 \right)d\] .

Substitute \[a=10\] and \[d=10\] in the equation.

\[\Rightarrow 100=10+\left( n-1 \right)10\]

\[\Rightarrow 100=10n\]

\[\Rightarrow n=10\]

The sum of first \[n\] terms of an arithmetic progression is given by the equation \[{{S}_{n}}=\frac{n}{2}\left[ 2a+\left( n-1 \right)d \right]\] .

Substitute the values of \[n\], \[a\] and \[d\] in the equation.

\[10+20+...+100=\frac{10}{2}\left[ 2\left( 10 \right)+\left( 10-1 \right)\left( 10 \right) \right]\]

\[=5\left[ 20+90 \right]\]

\[=5\left( 110 \right)\]

\[=550\]

Therefore, 

\[2250+1050-550=3050\]

is the required sum.

Therefore, \[3050\] is the sum of the integers from \[1\] to \[100\], which are divisible by \[2\]or \[5\].

6. Find the sum of all two-digit numbers which when divided by \[4\] , yields \[1\]  as remainder.

Ans:

\[13,17,...97\] are the two-digit numbers which when divided by \[4\] yields \[1\] as remainder. An A.P. is formed by this series.

The first term of the A.P. is \[a=13\], the last term is \[{{a}_{n}}=97\]  and the common difference is \[d=4\] .

The \[{{n}^{th}}\] term of the A.P. is given by the equation \[{{a}_{n}}=a+\left( n-1 \right)d\] .

Substitute \[a=13\] and \[d=4\] in the equation.

\[\Rightarrow 97=13+\left( n-1 \right)4\]

\[\Rightarrow 4\left( n-1 \right)=84\]

\[\Rightarrow n-1=21\]

\[\Rightarrow n=22\]

The sum of first \[n\] terms of an arithmetic progression is given by the equation \[{{S}_{n}}=\frac{n}{2}\left[ 2a+\left( n-1 \right)d \right]\] .

Substitute the values of \[n\], \[a\] and \[d\] in the equation.

\[{{S}_{22}}=\frac{22}{2}\left[ 2\left( 13 \right)+\left( 22-1 \right)\left( 4 \right) \right]\]

\[=11\left[ 26+84 \right]\]

\[=1210\]

Therefore, \[1210\] is the sum of all two-digit numbers which when divided by \[4\], yields \[1\] as remainder.

7. If is a function satisfying \[f\left( x+y \right)=f\left( x \right).f\left( y \right)\] for all \[x,y\in N\] , such that \[f\left( 1 \right)=3\] and \[\sum\limits_{x=1}^{n}{f\left( x \right)=120}\] find the value of \[n\].

Ans:

According to the given conditions in the question,

\[f\left( x+y \right)=f\left( x \right)\times f\left( y \right)\] for all \[x,y,\in N\]

\[f\left( 1 \right)=3\]

Let \[x=y=1\].

Then,

\[f\left( 1+1 \right)=f\left( 1+2 \right)=f\left( 1 \right)f\left( 2 \right)=3\times 3=9\]

We can also write 

\[f\left( 1+1+1 \right)=f\left( 3 \right)=f\left( 1+2 \right)=f\left( 1 \right)f\left( 2 \right)=3\times 9=27\]

\[f\left( 4 \right)=f\left( 1+4 \right)=f\left( 1 \right)f\left( 3 \right)=3\times 27=81\]

Both the first term and common ratio of \[f\left( 1 \right),f\left( 2 \right),f\left( 3 \right),...,\]that is \[3,9,27,...,\] that forms s G.P. is equal to \[3\]

We know that, \[{{S}_{n}}=\frac{a\left( {{r}^{n}}-1 \right)}{r-1}\]

Given that, \[\sum\limits_{k=1}^{n}{f}\left( x \right)=120\] 

Then,

\[120=\frac{3\left( {{3}^{n}}-1 \right)}{3-1}\]

\[\Rightarrow 120=\frac{3}{2}\left( {{3}^{n}}-1 \right)\]

\[\Rightarrow {{3}^{n}}-1=80\] 

\[\Rightarrow {{3}^{n}}=80={{3}^{4}}\]

\[\Rightarrow {{3}^{n}}-1=80\]

\[n=4\]

Therefore, \[4\] is the value of \[n\].

8. The sum of some terms of G.P. is \[315\] whose first term and the common ratio are \[5\] and \[2\], respectively. Find the last term and the number of terms.

Ans:

Let \[315\] be the sum of \[n\] terms of the G.P.

We know that, \[{{S}_{n}}=\frac{a\left( {{r}^{n}}-1 \right)}{r-1}\]

The first term \[a\] of the A.P. is \[5\] and the common difference \[r\] is \[2\].

Substitute the values of \[a\] and \[r\] in the equation

\[315=\frac{5\left( {{2}^{n}}-1 \right)}{2-1}\]

\[\Rightarrow {{2}^{n}}-1=63\]

\[\Rightarrow {{2}^{n}}=63={{\left( 2 \right)}^{2}}\]

\[\Rightarrow n=6\]

Therefore, the \[{{6}^{th}}\] term is the last term of the G.P.

 \[{{6}^{th}}\]term \[=a{{r}^{6-1}}=\left( 5 \right){{\left( 2 \right)}^{5}}=\left( 5 \right)\left( 32 \right)=160\]

Therefore, \[160\] is the last term of the G.P  and the number of terms is \[6\]. 

9. The first term of a G.P. is \[1\] . The sum of the third term and fifth term is \[90\]. Find the common ratio of G.P.

Ans:

Let the first term of the G.P. be \[a\] and the common ratio be \[r\] .

Then, \[a=1\]

\[{{a}_{3}}=a{{r}^{2}}={{r}^{2}}\]

\[{{a}_{5}}=a{{r}^{4}}={{r}^{4}}\]

Therefore,

\[{{r}^{2}}+{{r}^{4}}=90\]

\[\Rightarrow {{r}^{4}}+{{r}^{2}}-90=0\]

\[\Rightarrow {{r}^{2}}=\frac{-1+\sqrt{1+360}}{2}\]

\[=\frac{-1+\sqrt{361}}{2}\]

\[=-10\] or \[9\]

\[\Rightarrow r=\pm 3\]

Therefore, \[\pm 3\] is the common ratio of the G.P. 

10. The sum of the three numbers in G.P. is \[56\]. If we subtract \[1,7,21\] from these numbers in that order, we obtain an arithmetic progression. Find the numbers.

Ans:

Let \[a,ar\] and \[a{{r}^{2}}\] be the three numbers in G.P.

According to the conditions given in the question,

\[a+ar+a{{r}^{2}}=56\]

\[\Rightarrow a\left( 1+r+{{r}^{2}} \right)=56\]            …(1)

An A.P. is formed by

\[a-1,ar-7,a{{r}^{2}}-21\]

Therefore,

\[\left( ar-7 \right)-\left( a-1 \right)=\left( a{{r}^{2}}-21 \right)-\left( ar-7 \right)b\]

\[\Rightarrow ar-a-6=a{{r}^{2}}-ar-14\]

\[\Rightarrow a{{r}^{2}}-2ar+a=8\]

\[\Rightarrow a{{r}^{2}}-ar-ar+a=8\]

\[\Rightarrow a\left( {{r}^{2}}+1-2r \right)=8\]

\[\Rightarrow a{{\left( {{r}^{2}}-1 \right)}^{2}}=8\]                   …(2)

Equating (1) and (2), we get

\[\Rightarrow 7\left( {{r}^{2}}-2r+1 \right)=1+r+{{r}^{2}}\]

\[\Rightarrow 7{{r}^{2}}-14r+7-1-r-{{r}^{2}}\]

\[\Rightarrow 6{{r}^{2}}-15r+6=0\]

\[\Rightarrow 6{{r}^{2}}-12r-3r+6=0\]

\[\Rightarrow 6\left( r-2 \right)-3\left( r-2 \right)=0\]

\[\Rightarrow \left( 6r-3 \right)\left( r-2 \right)=0\]

Then,\[8,16\] and \[32\] are the three numbers when \[r=2\]  and \[32,16\] and \[8\] are the numbers when \[r=\frac{1}{2}\].  

Therefore, \[8,16\] and \[32\] are the three required numbers in either case. 

11.A G.P. consists of an even number of terms. If the sum of all the terms is \[5\]  times the sum of terms occupying odd places, then find its common ratio.

Ans: 

Let \[{{T}_{1}},{{T}_{2}},{{T}_{3}},{{T}_{4}},...{{T}_{2n}}\] be the G.P.

\[2n\] is the number of terms.

According to the conditions given in the question,

\[{{T}_{1}}+{{T}_{2}}+{{T}_{3}}+...+{{T}_{2n}}=5\left[ {{T}_{1}}+{{T}_{3}}+...+{{T}_{2n-1}} \right]\]

\[\Rightarrow {{T}_{1}}+{{T}_{2}}+{{T}_{3}}+...+{{T}_{2n}}-5\left[ {{T}_{1}}+{{T}_{3}}+...+{{T}_{2n-1}} \right]=0\]

\[\Rightarrow {{T}_{2}}+{{T}_{4}}+...+{{T}_{2n}}=4\left[ {{T}_{1}}+{{T}_{3}}+...+{{T}_{2n-1}} \right]\]

Let \[a,ar,a{{r}^{2}},a{{r}^{3}}\] be the G.P.

Therefore,

\[\frac{ar\left( {{r}^{n}}-1 \right)}{r-1}=\frac{4\times a\left( {{r}^{n}}-1 \right)}{r-1}\]

\[\Rightarrow ar=4a\]

\[\Rightarrow r=4\]

Therefore, \[4\] is the common ratio of the G.P. 

12. The sum of the first four terms of an A.P. is \[56\] . The sum of the last four terms is \[112\]. If its first term is \[11\], then find the number of terms.

Ans:

Let \[a,a+d,a+2d,a+3d...a+\left( n-2 \right)d,a+\left( n-2 \right)d\] be the A.P.

\[a\left( a+d \right)+\left( a+2d \right)+\left( a+3d \right)=4a+6d\] is the sum of the first four terms.

\[\left[ a+\left( n-4 \right)d \right]+\left[ a+\left( n-3 \right)d \right]+\left[ a+\left( n-2 \right)d \right]+\left[ a+\left( n-1 \right)d \right]=4a+\left( 4n-10 \right)d\] 

is the sum of the last four terms.

According to the conditions given in the question,

\[4a+6d=56\]

It is given that \[a=11\], then 

\[\Rightarrow 4\left( 11 \right)+6d=56\]

\[\Rightarrow 6d=12\]

\[\Rightarrow d=2\]

Therefore,

\[4a+\left( 4n-10 \right)d=112\]

\[\Rightarrow 4\left( 11 \right)+\left( 4n-10 \right)2=112\]

\[\Rightarrow \left( 4n-10 \right)2=68\]

\[\Rightarrow 4n-10=34\]

\[\Rightarrow 4n=44\]

\[\Rightarrow n=11\]

Therefore, \[11\] is the number of terms of the A.P.  

13: If \[\frac{a+bx}{a-bx}=\frac{b+cx}{b-cx}=\frac{c+dx}{c-dx}\left( x\ne 0 \right)\] then show that \[a,b,c\] and \[d\] are in G.P.

Ans:

Given ,

\[\frac{a+bx}{a-bx}=\frac{b+cx}{b-cx}\]

\[\Rightarrow \left( a+bx \right)\left( b-cx \right)=\left( b+cx \right)\left( a-bx \right)\]

\[\Rightarrow ab-acx+{{b}^{2}}x-bc{{x}^{2}}=ab-{{b}^{2}}x+-acx-bc{{x}^{2}}\]

\[\Rightarrow 2{{b}^{2}}x=2acx\]

\[\Rightarrow {{b}^{2}}=ac\]

\[\Rightarrow \frac{b}{a}=\frac{c}{b}\]

It is also given that,

\[\frac{b+cx}{b-cx}=\frac{c+dx}{c-dx}\]

\[\Rightarrow \left( b+cx \right)\left( c-dx \right)=\left( b-cx \right)\left( c+dx \right)\]

\[\Rightarrow bc-bdx+{{c}^{2}}x-cd{{x}^{2}}=bc+bdx-{{c}^{2}}x-cd{{x}^{2}}\]

\[\Rightarrow 2{{c}^{2}}x=2bdx\]

\[\Rightarrow {{c}^{2}}=bd\]

\[\Rightarrow \frac{c}{d}=\frac{d}{c}\]

Equating both the results, we get

\[\frac{b}{a}=\frac{c}{b}=\frac{d}{b}\]

Therefore, it is proved that \[a,b,c\] and \[d\] are in G.P.

14. Let \[S\] be the sum, \[P\] the product and \[R\] the sum of reciprocals of terms in a G.P. Prove that \[{{P}^{2}}{{R}^{n}}={{S}^{n}}\].

Ans:

Let \[a,ar,a{{r}^{2}},a{{r}^{3}}...a{{r}^{n-1}}\] be the G.P.

According to the conditions given in the question,

\[S=\frac{a\left( {{r}^{n}}-1 \right)}{r-1}\] 

\[P={{a}^{n}}\times {{r}^{1+2+...+n-1}}\] 

Since the sum of first \[n\] natural numbers is \[n\frac{\left( n+1 \right)}{2}\]

\[\Rightarrow P={{a}^{n}}{{r}^{\frac{n\left( n-1 \right)}{2}}}\] 

\[R=\frac{1}{a}+\frac{1}{ar}+...+\frac{1}{a{{r}^{n-1}}}\]

\[=\frac{{{r}^{n-1}}+{{r}^{n-2}}+...r+1}{a{{r}^{n-1}}}\]

Since \[1,r,...{{r}^{n-1}}\]forms a G.P.,

\[\Rightarrow R=\frac{1\left( {{r}^{n}}-1 \right)}{\left( r-1 \right)}\times \frac{1}{a{{r}^{n-1}}}\]            

\[=\frac{{{r}^{n}}-1}{a{{r}^{n-1}}\left( r-1 \right)}\]

Then,

\[{{P}^{2}}{{R}^{n}}={{a}^{2n}}{{r}^{n\left( n-1 \right)}}\frac{{{\left( {{r}^{n}}-1 \right)}^{n}}}{{{a}^{n}}{{r}^{n\left( n-1 \right)}}{{\left( r-1 \right)}^{n}}}\]

\[=\frac{{{a}^{n}}{{\left( {{r}^{n}}-1 \right)}^{n}}}{{{\left( r-1 \right)}^{n}}}\]

\[={{\left[ \frac{a\left( {{r}^{n}}-1 \right)}{\left( r-1 \right)} \right]}^{n}}\]

\[={{S}^{n}}\]

Therefore, \[{{P}^{2}}{{R}^{n}}={{S}^{n}}\].

15. The \[{{p}^{th}}\],\[{{q}^{th}}\] and \[{{r}^{th}}\] terms of an A.P. are \[a,b,c\] respectively.  Show that 

\[\left( q-r \right)a+\left( r-p \right)b+\left( p-q \right)c=0\].

Ans:

Let the first term of the A.P. be \[t\] and the common difference be \[d\] . 

\[{{a}_{n}}=t+\left( n-1 \right)d\] is the equation of the \[{{n}^{th}}\] term of the A.P.

According to the conditions given in the question,

\[{{a}_{p}}=t+\left( p-1 \right)d=a\]

\[{{a}_{q}}=t+\left( q-1 \right)d=b\]

\[{{a}_{r}}=t+\left( r-1 \right)d=c\]

Subtract \[{{a}_{q}}\] from \[{{a}_{p}}\].

\[\left( p-1-q+1 \right)d=a-b\]

\[\Rightarrow \left( p-q \right)d=a-b\]

\[\Rightarrow d=\frac{a-b}{p-q}\]

Subtract \[{{a}_{r}}\] from \[{{a}_{q}}\].

\[\left( q-1-r+1 \right)d=b-c\]

\[\Rightarrow \left( q-r \right)d=b-c\]

\[\Rightarrow d=\frac{b-c}{q-r}\]

Equating both the values of \[d\] obtained, we get

\[\frac{a-b}{p-q}=\frac{b-c}{q-r}\]

\[\Rightarrow \left( a-b \right)\left( q-r \right)=\left( b-c \right)\left( p-q \right)\]

\[\Rightarrow aq-bq-ar+br=bp-bq-cp+cq\]

\[\Rightarrow bp-cp+cq-aq+ar-br=0\]

By rearranging terms we get

\[\Rightarrow \left( -aq+ar \right)+\left( bp-br \right)+\left( -cp+cq \right)=0\]

\[\Rightarrow -a\left( q-r \right)-b\left( r-p \right)-c\left( p-q \right)=0\]

\[\Rightarrow a\left( q-r \right)+b\left( r-p \right)+c\left( p-q \right)=0\]

Therefore, \[\left( q-r \right)a+\left( r-p \right)b+\left( p-q \right)c=0\] is proved.

16. If \[a\left( \frac{1}{b}+\frac{1}{c} \right),b\left( \frac{1}{c}+\frac{1}{a} \right),c\left( \frac{1}{a}+\frac{1}{b} \right)\] are in A.P., prove that \[a,b,c\] are in A.P.

Ans:

Given \[a\left( \frac{1}{b}+\frac{1}{c} \right),b\left( \frac{1}{c}+\frac{1}{a} \right),c\left( \frac{1}{a}+\frac{1}{b} \right)\] are in A.P.

Therefore,

\[b\left( \frac{1}{c}+\frac{1}{a} \right)-a\left( \frac{1}{b}+\frac{1}{c} \right)=c\left( \frac{1}{a}+\frac{1}{b} \right)-b\left( \frac{1}{c}+\frac{1}{a} \right)\]

\[\Rightarrow \frac{b\left( a+c \right)}{ac}-\frac{a\left( b+c \right)}{bc}=\frac{c\left( a+b \right)}{ab}-\frac{b\left( a+c \right)}{ac}\]

\[\Rightarrow \frac{{{b}^{2}}a+{{b}^{2}}c-{{a}^{2}}b-{{a}^{2}}c}{abc}=\frac{{{c}^{2}}a+{{c}^{2}}b-{{b}^{2}}a-{{b}^{2}}c}{abc}\]

\[\Rightarrow {{b}^{2}}a+{{b}^{2}}c-{{a}^{2}}b-{{a}^{2}}c={{c}^{2}}a+{{c}^{2}}b-{{b}^{2}}a-{{b}^{2}}c\]

\[\Rightarrow ab\left( b-a \right)+c\left( {{b}^{2}}-{{a}^{2}} \right)=a\left( {{c}^{2}}-{{b}^{2}} \right)bc\left( c-b \right)\]

\[\Rightarrow ab\left( b-a \right)+c\left( b-a \right)\left( b+a \right)=a\left( c-b \right)\left( c+b \right)+bc\left( c-b \right)\]

\[\Rightarrow \left( b-a \right)\left( ab+cb+ca \right)=\left( c-b \right)\left( ac+ab+bc \right)\]

\[\Rightarrow b-a=c-b\]

Therefore, \[a,b\] and \[c\] are in A.P.

17. If \[a,b,c,d\] are in G.P., prove that \[\left( {{a}^{n}}+{{b}^{n}} \right),\left( {{b}^{n}}+{{c}^{n}} \right),\left( {{c}^{n}}+{{d}^{n}} \right)\] are in G.P

Ans: 

Given: 

\[a,b,c\] and \[d\] are in G.P.

Therefore,

\[{{b}^{2}}=ac\]

\[{{c}^{2}}=bd\]

\[ad=bc\]

To prove: 

\[\left( {{a}^{n}}+{{b}^{n}} \right),\left( {{b}^{n}}+{{c}^{n}} \right),\left( {{c}^{n}}+{{d}^{n}} \right)\] are in G.P.

That is, \[{{\left( {{b}^{n}}+{{c}^{n}} \right)}^{2}}=\left( {{a}^{n}}+{{b}^{n}} \right),\left( {{c}^{n}}+{{d}^{n}} \right)\]

Then, 

L.H.S \[={{\left( {{b}^{n}}+{{c}^{n}} \right)}^{2}}\]

\[={{b}^{2n}}+2{{b}^{n}}{{c}^{n}}+{{c}^{2n}}\]

\[={{\left( {{b}^{2}} \right)}^{n}}+2{{b}^{n}}{{c}^{n}}+{{\left( {{c}^{2}} \right)}^{n}}\]

\[={{\left( ac \right)}^{n}}+2{{b}^{n}}{{c}^{n}}+{{\left( bd \right)}^{n}}\]

\[={{a}^{n}}{{c}^{n}}+{{b}^{n}}{{c}^{n}}+{{b}^{n}}{{c}^{n}}+{{b}^{n}}{{d}^{n}}\]

\[={{a}^{n}}{{c}^{n}}+{{b}^{n}}{{c}^{n}}+{{a}^{n}}{{d}^{n}}+{{b}^{n}}{{d}^{n}}\]

\[={{c}^{n}}\left( {{a}^{n}}+{{b}^{n}} \right)+{{d}^{n}}\left( {{a}^{n}}+{{b}^{n}} \right)\]

\[=\left( {{a}^{n}}+{{b}^{n}} \right)\left( {{a}^{n}}+{{d}^{n}} \right)\]

\[=\]R.H.S

Therefore,

\[{{\left( {{b}^{n}}+{{c}^{n}} \right)}^{2}}=\left( {{a}^{n}}+{{b}^{n}} \right)\left( {{c}^{n}}+{{d}^{n}} \right)\]

Therefore, \[\left( {{b}^{n}}+{{c}^{n}} \right),\left( {{b}^{n}}+{{c}^{n}} \right)\] and \[\left( {{c}^{n}}+{{d}^{n}} \right)\] are in G.P.

18. If \[a\] and \[b\] are the roots of \[{{x}^{2}}-3x+p=0\] and \[c,d\] are roots of \[{{x}^{2}}-12x+q=0\], where \[a,b,c,d\] form a G.P. Prove that \[\left( q+p \right):\left( q-p \right)=17:15\] . 

Ans:

Given: \[a\] and \[b\] are the roots of \[{{x}^{2}}-3x+p=0\].

Therefore,

\[a+b=3\] and \[ab=p\]    …(1)      

We also know that \[c\] and \[d\] are the roots of \[{{x}^{2}}-12x+q=0\].

Therefore,

\[c+d=12\] and \[cd=q\]   …(2)      

Also, \[a,b,c,d\] are in G.P.

Let us take \[a=x,b=xr,c=x{{r}^{2}}\] and \[d=x{{r}^{3}}\].

We get from (1) and (2) that,

\[x+xr=3\]

\[\Rightarrow x\left( 1+r \right)=3\]

Also,

\[x{{r}^{2}}+x{{r}^{3}}=12\]

\[\Rightarrow x{{r}^{2}}+\left( 1+r \right)=12\]

Divide both the equations obtained.

\[\frac{x{{r}^{2}}\left( 1+r \right)}{x\left( 1+r \right)}=\frac{12}{3}\]

\[\Rightarrow {{r}^{2}}=4\]

\[\Rightarrow r=\pm 2\]

\[x=\frac{3}{1+2}=\frac{3}{3}=1\], when \[r=2\] and

\[x=\frac{3}{1-2}=\frac{3}{-1}=-3\], when \[r=-2\].

Case I:

\[ab={{x}^{2}}r=2\], \[cd={{x}^{2}}{{r}^{5}}=32\] when \[r=2\] and \[x=1\] .

Therefore, 

\[\frac{q+p}{q-p}=\frac{32+2}{32-2}=\frac{34}{30}=\frac{17}{15}\]

\[\Rightarrow \left( q+p \right):\left( q-p \right)=17:15\]

Case II:

\[ab={{x}^{2}}r=18\], \[cd={{x}^{2}}{{r}^{5}}=-288\] when \[r=-2\] and \[x=-3\] .

Therefore,

\[\frac{q+p}{q-p}=\frac{-288-18}{-288+18}=\frac{-306}{-270}=\frac{17}{15}\]

\[\Rightarrow \left( q+p \right):\left( q-p \right)=17:15\]

Therefore, it is proved that \[\left( q+p \right):\left( q-p \right)=17:15\]as we obtain the same for both the cases. 

19. The ratio of the A.M and G.M. of two positive numbers \[a\] and \[b\] is \[m:n\]. Show that \[a:b=\left( m+\sqrt{{{m}^{2}}-{{n}^{2}}} \right):\left( m-\sqrt{{{m}^{2}}-{{n}^{2}}} \right)\] .

Ans:

Let \[a\] and \[b\] be the two numbers.

The arithmetic mean, A.M \[=\frac{a+b}{2}\] and the geometric mean, G.M \[=\sqrt{ab}\]

According to the conditions given in the question,

\[\frac{a+b}{2\sqrt{ab}}=\frac{m}{n}\]

\[\Rightarrow \frac{{{\left( a+b \right)}^{2}}}{4\left( ab \right)}=\frac{{{m}^{2}}}{{{n}^{2}}}\]

\[\Rightarrow \left( a+b \right)=\frac{4ab{{m}^{2}}}{{{n}^{2}}}\]

\[\Rightarrow \left( a+b \right)=\frac{2\sqrt{ab}m}{n}\]                         …(1)

Using the above equation in the identity \[{{\left( a-b \right)}^{2}}={{\left( a+b \right)}^{2}}-4ab\] , we obtain

\[{{\left( a-b \right)}^{2}}=\frac{4ab{{m}^{2}}}{{{n}^{2}}}-4ab=\frac{4ab\left( {{m}^{2}}-{{n}^{2}} \right)}{{{n}^{2}}}\]

\[\Rightarrow \left( a-b \right)=\frac{2\sqrt{ab}\sqrt{{{m}^{2}}-{{n}^{2}}}}{n}\]             …(2)

Add equation (1) and (2)

\[2a=\frac{2\sqrt{ab}}{n}\left( m+\sqrt{{{m}^{2}}-{{n}^{2}}} \right)\]

\[\Rightarrow a=\frac{\sqrt{ab}}{n}\left( m+\sqrt{{{m}^{2}}-{{n}^{2}}} \right)\]

Substitute in (1) the value of \[a\].

\[b=\frac{2\sqrt{ab}}{n}m-\frac{\sqrt{ab}}{n}\left( m+\sqrt{{{m}^{2}}-{{n}^{2}}} \right)\]

\[=\frac{\sqrt{ab}}{n}m-\frac{\sqrt{ab}}{n}\sqrt{{{m}^{2}}-{{n}^{2}}}\]

\[=\frac{\sqrt{ab}}{n}\left( m-\sqrt{{{m}^{2}}-{{n}^{2}}} \right)\]

Therefore,

\[a:b=\frac{a}{b}=\frac{\frac{\sqrt{ab}}{n}\left( m+\sqrt{{{m}^{2}}-{{n}^{2}}} \right)}{\frac{\sqrt{ab}}{n}\left( m-\sqrt{{{m}^{2}}-{{n}^{2}}} \right)}=\frac{\left( m+\sqrt{{{m}^{2}}-{{n}^{2}}} \right)}{\left( m-\sqrt{{{m}^{2}}-{{n}^{2}}} \right)}\]

Therefore, it is proved that \[a:b=\left( m+\sqrt{{{m}^{2}}-{{n}^{2}}} \right):\left( m-\sqrt{{{m}^{2}}-{{n}^{2}}} \right)\].

20. If \[a,b,c\] are in A.P; \[b,c,d\] are in G.P. and \[\frac{1}{c},\frac{1}{d},\frac{1}{e}\] are in A.P. Prove that \[a,c,e\] are in G.P.

Ans:

Given \[a,b,c\] are in A.P.

Therefore, \[b-a=c-b\]

\[\Rightarrow 2b=a+c\]

\[\Rightarrow b=\frac{a+c}{2}\]       …(1)

It is given \[b,c,d\] are in G.P.

Therefore, \[{{c}^{2}}=bd\]

\[\Rightarrow d=\frac{{{c}^{2}}}{b}\]           …(2)

Also, \[\frac{1}{c},\frac{1}{d},\frac{1}{e}\] are in A.P.

\[\frac{1}{d}-\frac{1}{c}=\frac{1}{e}-\frac{1}{d}\]

\[\Rightarrow \frac{2}{d}=\frac{1}{c}+\frac{1}{e}\]       …(3)

We have to prove that \[a,c,e\] are in G.P.

That is, \[{{c}^{2}}=ae\].

Substitute (1) and (2) in (3).

\[\frac{2b}{{{c}^{2}}}=\frac{1}{c}+\frac{1}{e}\]

\[\Rightarrow \frac{2\left( a+c \right)}{{{c}^{2}}}=\frac{1}{c}+\frac{1}{e}\]

\[\Rightarrow \frac{a+c}{{{c}^{2}}}=\frac{e+c}{ce}\]

\[\Rightarrow \frac{a+c}{c}=\frac{e+c}{e}\]

\[\Rightarrow \left( a+c \right)e=\left( e+c \right)c\]

\[\Rightarrow ae+ce=ec+{{c}^{2}}\]

\[\Rightarrow {{c}^{2}}=ae\]

Therefore, it is proved that \[a,c,e\] are in G.P.

21. Find the sum of the following series up to \[n\] terms:

1. \[5+55+555+...\]

Ans:

Let \[{{S}_{n}}=5+55+555...\] to \[n\] terms.

\[=\frac{5}{9}\](9+99+999+... to n terms.)

\[=\frac{5}{9}(( 10-1 )+( {{10}^{2}}-1 )+( {{10}^{3}}-1)+...\]to n terms)

\[=\frac{5}{9}((10+{{10}^{2}}+{{10}^{3}}...\]to n terms)-(1+1+ to n terms))

\[=\frac{5}{9}\left[ \frac{10\left( {{10}^{n}}-1 \right)}{10-1}-n \right]\]

\[=\frac{5}{9}\left[ \frac{10\left( {{10}^{n}}-1 \right)}{9}-n \right]\]

\[=\frac{50}{81}\left( {{10}^{n}}-1 \right)-\frac{5n}{9}\]

Therefore, the sum of \[n\] terms of the given series is \[\frac{50}{81}\left( {{10}^{n}}-1 \right)-\frac{5n}{9}\] .

2. \[.6+.66+.666.+...\]

Ans:

Let \[{{S}_{n}}=0.6+0.66+0.666+\] to \[n\] terms.

\[=6\] (0.1+0.11+0.111+... to \[n\] terms)

\[=\frac{6}{9}\] (0.9+0.99+0.999+... to \[n\] terms)

\[=\frac{6}{9} (\left( 1-\frac{1}{10} \right)+\left( 1-\frac{1}{{{10}^{2}}} \right)+\left( 1-\frac{1}{{{10}^{3}}} \right))+... \]to n terms

\[=\frac{2}{3}\]((\[1+1+...\] to \[n\] terms)\[-\] \[\frac{1}{10}\] (\[1+\frac{1}{10}+\frac{1}{{{10}^{2}}}\] to \[n\] terms))

\[=\frac{2}{3}( n-\frac{1}{10}\left( \frac{1-{{\left( \frac{1}{10} \right)}^{n}}}{1-\frac{1}{10}} \right) )\]

\[=\frac{2}{3}n-\frac{2}{30}\times \frac{10}{9}\left( 1-{{10}^{-n}} \right)\]

\[=\frac{2}{3}n-\frac{2}{27}\left( 1-{{10}^{-n}} \right)\]

Therefore, the sum of \[n\] terms of the given series is \[\frac{2}{3}n-\frac{2}{27}\left( 1-{{10}^{-n}} \right)\] .

22. Find the \[{{20}^{th}}\] term of the series \[2\times 4+4\times 6+6\times 8+...+n\] terms.

Ans:

\[2\times 4+4\times 6+6\times 8+...+n\] is the given series,

Therefore the \[{{n}^{th}}\] term \[{{a}_{n}}=2n\times \left( 2n+2 \right)=4{{n}^{2}}+4n\]

Then,

\[{{a}_{20}}=4{{\left( 20 \right)}^{2}}+4\left( 20 \right)\]

\[=4\left( 400 \right)+80\]

\[=1600+80\]

\[=1680\]

Therefore, \[1680\] is the \[{{20}^{th}}\] term of the series. 

23. Find the sum of the first \[n\] terms of the series: \[3+7+13+21+31+...\] 

Ans:

\[3+7+13+21+31+...\] is the given series.

\[S=3+7+13+21+31+...+{{a}_{n-1}}+{{a}_{n}}\]

\[S=3+7+13+21+...+{{a}_{n-2}}+{{a}_{n-1}}+{{a}_{n}}\]

Subtract both the equations.

\[S-S=\left[ 3+\left( 7+13+21+31+...+{{a}_{n-1}}+{{a}_{n}} \right)+ \right]-\left[ \left( 3+7+13+21+31+...+{{a}_{n-1}} \right)+{{a}_{n}} \right]\]

\[\Rightarrow S-S=3+\left[ \left( 7-3 \right)+\left( 13-7 \right)+\left( 21-13 \right)+...+\left( {{a}_{n}}-{{a}_{n-1}} \right) \right]-{{a}_{n}}\]

\[\Rightarrow 0=3+\left[ 4+6+8+...+\left( n-1 \right)terms \right]-{{a}_{n}}\]

\[\Rightarrow {{a}_{n}}=3+\left[ 4+6+8+...+\left( n-1 \right)terms \right]\]

\[\Rightarrow {{a}_{n}}=3+\left( \frac{n-1}{2} \right)\left[ 2\times 4+\left( n-1-1 \right)2 \right]\]

\[=3+\left( \frac{n-1}{2} \right)\left[ 8+\left( n-2 \right)2 \right]\]

\[=3+\left( \frac{n-1}{2} \right)\left( 2n+4 \right)\]

\[=3+\left( n-1 \right)\left( n+2 \right)\]

\[=3+\left( {{n}^{2}}+n-2 \right)\]

\[={{n}^{2}}+n+1\]

Therefore, 

\[\sum\limits_{k=1}^{n}{{{a}_{k}}}=\sum\limits_{k=1}^{n}{{{k}^{2}}}+\sum\limits_{k=1}^{n}{k}+\sum\limits_{k=1}^{n}{1}\]

\[=\frac{n\left( n+1 \right)\left( 2n+1 \right)}{6}+\frac{n\left( n+1 \right)}{2}+n\]

\[=n\left[ \frac{\left( n+1 \right)\left( 2n+1 \right)+3\left( n+1 \right)+6}{6} \right]\]

\[=n\left[ \frac{2{{n}^{2}}+3n+1+3n+3+36}{6} \right]\]

\[=n\left[ \frac{2{{n}^{2}}+6n+10}{6} \right]\]

\[=\frac{n}{3}\left[ {{n}^{2}}+3n+5 \right]\]

Therefore, the sum of \[n\] terms of the given series is \[\frac{n}{3}\left[ {{n}^{2}}+3n+5 \right]\].

24. If \[{{S}_{1}},{{S}_{2}},{{S}_{3}}\] are the sum of first \[n\] natural numbers, their squares and their cubes, respectively, show that \[9S_{2}^{2}={{S}_{3}}\left( 1+8{{S}_{1}} \right)\].

Ans:

According to the conditions given in the question,

\[{{S}_{1}}=\frac{n\left( n+1 \right)}{2}\]

\[{{S}_{3}}=\frac{{{n}^{2}}{{\left( n+1 \right)}^{2}}}{4}\]

Therefore,

\[{{S}_{3}}\left( 1+8{{S}_{1}} \right)=\frac{{{n}^{2}}{{\left( n+1 \right)}^{2}}}{4}\left[ 1+\frac{8n\left( n+1 \right)}{2} \right]\]

\[=\frac{{{n}^{2}}{{\left( n+1 \right)}^{2}}}{4}\left[ 1+4{{n}^{2}}+4n \right]\]

\[=\frac{{{n}^{2}}{{\left( n+1 \right)}^{2}}}{4}{{\left( 2n+1 \right)}^{2}}\]

\[=\frac{{{\left[ n\left( n+1 \right)\left( 2n+1 \right) \right]}^{2}}}{4}\]        …(1)

And. \[9S_{2}^{2}=9\frac{{{\left[ n\left( n+1 \right)\left( 2n+1 \right) \right]}^{2}}}{{{\left( 6 \right)}^{2}}}\]

\[=\frac{9}{36}{{\left[ n\left( n+1 \right)\left( 2n+1 \right) \right]}^{2}}\]

\[=\frac{{{\left[ n\left( n+1 \right)\left( 2n+1 \right) \right]}^{2}}}{4}\]        …(2)

Therefore, we get \[9S_{2}^{2}={{S}_{3}}\left( 1+8{{S}_{1}} \right)\] from (1) and (2).

25. Find the sum of the following series up to \[n\] terms:

\[\frac{{{1}^{3}}}{1}+\frac{{{1}^{3}}+{{2}^{3}}}{1+3}+\frac{{{1}^{3}}+{{2}^{3}}+{{3}^{3}}}{1+3+5}+...\]

Ans:

\[\frac{{{1}^{3}}+{{2}^{3}}+{{3}^{3}}+...+{{n}^{3}}}{1+3+5+...+\left( 2n-1 \right)}=\frac{{{\left[ \frac{n\left( n+1 \right)}{2} \right]}^{2}}}{1+3+5+...+\left( 2n-1 \right)}\] is the \[{{n}^{th}}\] term of the given series.

With first term \[a\] , last term \[\left( 2n-1 \right)\] and number of terms as \[n\] then \[1,3,5...\left( 2n-1 \right)\] is an A.P.

Therefore, 

\[1+3+5+...+\left( 2n-1 \right)=\frac{n}{2}\left[ 2\times 1+\left( n-1 \right)2 \right]={{n}^{2}}\]

Also,

\[{{a}_{n}}=\frac{{{n}^{2}}{{\left( n+1 \right)}^{2}}}{4{{n}^{2}}}=\frac{{{\left( n+1 \right)}^{2}}}{4}=\frac{1}{4}{{n}^{2}}+\frac{1}{2}n+\frac{1}{4}\]

And,

\[{{S}_{n}}=\sum\limits_{k=1}^{n}{{{a}_{k}}}=\sum\limits_{k=1}^{n}{\left( \frac{1}{4}{{K}^{2}}+\frac{1}{2}K+\frac{1}{4} \right)}\]

\[=\frac{1}{4}\frac{n\left( n+1 \right)\left( 2n+1 \right)}{6}+\frac{1}{2}\frac{n\left( n+1 \right)}{2}+\frac{1}{4}n\]

\[=\frac{n\left[ \left( n+1 \right)\left( 2n+1 \right)+6\left( n+1 \right)+6 \right]}{24}\]

\[=\frac{n\left[ 2{{n}^{2}}+3n+1+6n+6+6 \right]}{24}\]

\[=\frac{n\left[ 2{{n}^{2}}+9n+13 \right]}{24}\]

Therefore, the sum of \[n\] terms of the given series \[\frac{n\left[ 2{{n}^{2}}+9n+13 \right]}{24}\] .

26. Show that \[\frac{1\times {{2}^{2}}+2\times {{3}^{2}}+...+n\times {{\left( n+1 \right)}^{2}}}{{{1}^{2}}\times 2+{{2}^{2}}\times 3+...+{{n}^{2}}\times \left( n+1 \right)}=\frac{3n+5}{3n+1}\] .

Ans:

\[n{{\left( n+1 \right)}^{2}}={{n}^{3}}+2{{n}^{2}}+n\] is the \[{{n}^{th}}\] term of the numerator.

\[{{n}^{2}}\left( n+1 \right)={{n}^{3}}+{{n}^{2}}\] is the \[{{n}^{th}}\] term of the denominator.

\[\frac{1\times {{2}^{2}}+2\times {{3}^{2}}+...+n\times {{\left( n+1 \right)}^{2}}}{{{1}^{2}}\times 2+{{2}^{2}}\times 3+...+{{n}^{2}}\times \left( n+1 \right)}=\frac{\sum\limits_{k=1}^{n}{{{a}_{k}}}}{\sum\limits_{k=1}^{n}{{{a}_{k}}}}=\frac{\sum\limits_{k=1}^{n}{\left( {{K}^{3}}+2{{K}^{2}}+K \right)}}{\sum\limits_{k=1}^{n}{\left( {{K}^{3}}+{{K}^{2}} \right)}}\]      …(1)

\[\sum\limits_{k=1}^{n}{\left( {{K}^{3}}+2{{K}^{2}}+K \right)}\]

\[=\frac{{{n}^{2}}{{\left( n+1 \right)}^{2}}}{4}+\frac{2n\left( n+1 \right)\left( 2n+1 \right)}{6}+\frac{n\left( n+1 \right)}{2}\]

\[=\frac{n\left( n+1 \right)}{2}\left[ \frac{n\left( n+1 \right)}{2}+\frac{2}{3}\left( 2n+1 \right)+1 \right]\]

\[=\frac{n\left( n+1 \right)}{2}\left[ \frac{3{{n}^{2}}+3n+8n+4+6}{6} \right]\]

\[=\frac{n\left( n+1 \right)}{12}\left[ 3{{n}^{2}}+11n+10 \right]\]

\[=\frac{n\left( n+1 \right)}{12}\left[ 3{{n}^{2}}+6n+5n+10 \right]\]

\[=\frac{n\left( n+1 \right)}{12}\left[ 3n\left( n+2 \right)+5\left( n+2 \right) \right]\]

\[=\frac{n\left( n+1 \right)\left( n+2 \right)\left( 3n+5 \right)}{12}\]                     …(2)

And,

\[\sum\limits_{k=1}^{n}{\left( {{K}^{3}}+{{K}^{2}} \right)}\]

\[=\frac{{{n}^{2}}{{\left( n+1 \right)}^{2}}}{4}+\frac{n\left( n+1 \right)\left( 2n+1 \right)}{6}\]

\[=\frac{n\left( n+1 \right)}{2}\left[ \frac{n\left( n+1 \right)}{2}+\frac{2n+1}{3} \right]\]

\[=\frac{n\left( n+1 \right)}{2}\left[ \frac{3{{n}^{2}}+3n+4n+2}{6} \right]\]

\[=\frac{n\left( n+1 \right)}{12}\left[ 3{{n}^{2}}+7n+2 \right]\]

\[=\frac{n\left( n+1 \right)}{12}\left[ 3{{n}^{2}}+6n+n+2 \right]\]

\[=\frac{n\left( n+1 \right)}{12}\left[ 3n\left( n+2 \right)+1\left( n+2 \right) \right]\]

\[=\frac{n\left( n+1 \right)\left( n+2 \right)\left( 3n+1 \right)}{12}\]                     …(3)

Substituting (2) and (3) in (1), we get

\[\frac{1\times {{2}^{2}}+2\times {{3}^{2}}+...+n\times {{\left( n+1 \right)}^{2}}}{{{1}^{2}}\times 2+{{2}^{2}}\times 3+...+{{n}^{2}}\times \left( n+1 \right)}=\frac{\frac{n\left( n+1 \right)\left( n+2 \right)\left( 3n+5 \right)}{12}}{\frac{n\left( n+1 \right)\left( n+2 \right)\left( 3n+1 \right)}{12}}\]

\[=\frac{n\left( n+1 \right)\left( n+2 \right)\left( 3n+5 \right)}{n\left( n+1 \right)\left( n+2 \right)\left( 3n+1 \right)}\]

\[=\frac{3n+5}{3n+1}\]

Therefore, \[\frac{1\times {{2}^{2}}+2\times {{3}^{2}}+...+n\times {{\left( n+1 \right)}^{2}}}{{{1}^{2}}\times 2+{{2}^{2}}\times 3+...+{{n}^{2}}\times \left( n+1 \right)}=\frac{3n+5}{3n+1}\] is proved.

27. A farmer buys a used tractor for Rs.\[12000\]. He pays Rs.\[6000\] cash and agrees to pay the balance in annual installments of Rs.\[500\] plus \[12%\] interest on the unpaid amount. How much will be the tractor cost him?

Ans:

It is given that Rs.\[6000\] is paid in cash by the farmer.

Therefore, the unpaid amount is given by

Rs.\[12000-\] Rs.\[6000=\]Rs.\[6000\]

According to the conditions given in the question, the interest to be paid annually by the farmer is 

\[12%\] of \[6000\] , \[12%\] of \[5500\] , \[12%\] of \[5000...12%\] of \[500\]

Therefore, the total interest to be paid by the farmer

\[=12%\] of \[6000+12%\] of \[5500+12%\] of \[5000+...+12%\] of \[500\]

\[=12%\] of \[\left( 6000+5500+5000+...+500 \right)\]

\[=12%\] of \[\left( 500+1000+1500+...+6000 \right)\]

With both the first term and common difference equal to \[500\], the series \[500,1000,1500...6000\] is an A.P.

Let \[n\] be the number of terms of the A.P. 

Therefore,

\[6000=500+\left( n-1 \right)500\]

\[\Rightarrow 1+\left( n-1 \right)=12\]

\[\Rightarrow n=12\]

Therefore, the sum of the given A.P.

\[=\frac{12}{2}\left[ 2\left( 500 \right)+\left( 12-1 \right)\left( 500 \right) \right]\]

\[=6\left[ 1000+5500 \right]\]

\[=6\left( 6500 \right)\]

\[=39000\]

Therefore, the total interest to be paid by the farmer

\[=12%\] of \[\left( 500+1000+1500+...+6000 \right)\]

\[=12%\] of Rs.\[39000\]

\[=\] Rs.\[4680\] 

Therefore, the total cost of tractor

\[=\](Rs.\[12000+\]Rs.\[4680\])

\[=\]Rs.\[16680\]

Therefore, the total cost of the tractor is Rs.\[16680\].

28. Shamshad Ali buys a scooter for Rs.\[22000\]. He pays Rs.\[4000\] cash and agrees to pay the balance in annual installment of Rs.\[1000\] plus \[10%\] interest on the unpaid amount. How much will the scooter cost him?

Ans:

It is given that for Rs.\[22000\] Shamshad Ali buys a scooter and Rs.\[4000\] is paid in cash.

Therefore, the unpaid amount is given by

Rs.\[22000-\] Rs.\[4000=\]Rs.\[18000\]

According to the conditions given in the question, the interest to be paid annually

is 

\[10%\] of \[18000\] , \[10%\] of \[17000\] , \[10%\] of \[16000...10%\] of \[1000\]

Therefore, the total interest to be paid by the farmer

\[=10%\] of \[18000+10%\] of \[17000+10%\] of \[16000+...+10%\] of \[1000\]

\[=10%\] of \[\left( 18000+17000+16000+...+1000 \right)\]

\[=10%\] of \[\left( 1000+2000+3000+...+18000 \right)\]

With both the first term and common difference equal to \[1000\], the series \[1000,2000,3000...18000\] is an A.P.

Let \[n\] be the number of terms of the A.P. 

Therefore,

\[18000=1000+\left( n-1 \right)1000\]

\[\Rightarrow 1+\left( n-1 \right)=18\]

\[\Rightarrow n=18\]

Therefore, the sum of the given A.P.

\[=\frac{18}{2}\left[ 2\left( 1000 \right)+\left( 18-1 \right)\left( 1000 \right) \right]\]

\[=9\left[ 2000+17000 \right]\]

\[=9\left( 19000 \right)\]

\[=171000\]

Therefore, the total interest to be paid

\[=10%\] of \[\left( 18000+17000+16000+...+1000 \right)\]

\[=10%\] of  Rs.\[171000\]

\[=\] Rs.\[17100\] 

Therefore, the total cost of scooter

\[=\](Rs.\[22000+\]Rs.\[17100\])

\[=\]Rs.\[39100\]

Therefore, the total cost of the scooter is Rs.\[39100\] .

29. A person writes a letter to four of his friends. He asks each one of them to copy the letter and mail to four different persons with instruction that they move the chain similarly. Assuming that the chain is not broken and that it costs \[50\] paise to mail one letter. Find the amount spent on the postage when \[{{8}^{th}}\] set of letter is mailed.

Ans:

\[4,{{4}^{2}},{{...4}^{8}}\] is the number of letters mailed and it forms a G.P.

The first term \[a=4\] , the common ratio \[r=4\] and the number of terms \[n=8\] of the G.P.

We know that the sum of \[n\] terms of a G.P. is 

\[{{S}_{n}}=\frac{a\left( {{r}^{n}}-1 \right)}{r-1}\]

Therefore,

\[{{S}_{8}}=\frac{4\left( {{4}^{8}}-1 \right)}{4-1}\]

\[=\frac{4\left( 65536-1 \right)}{3}\]

\[=\frac{4\left( 65535 \right)}{3}\]

\[=4\left( 21845 \right)\]

\[=87380\]

\[50\] paisa is the cost to mail one letter.

Therefore,

Cost of mailing \[87380\] letters \[=\] Rs.\[87380\times \frac{50}{100}\] \[=\] Rs.\[43690\]

Therefore, Rs.\[43690\] is the amount spent when \[{{8}^{th}}\] set of letter is mailed.

30. A man deposited Rs.\[10000\] in a bank at the rate of \[5%\] simple interest annually. Find the amount in \[{{15}^{th}}\] year since he deposited the amount and also calculate the total amount after \[20\] years.

Ans:

Rs.\[10000\] is deposited by the man in a bank at the rate of \[5%\] simple interest annually

\[=\frac{5}{100}\times \]Rs.\[10000=\]Rs.\[500\]

Therefore,

\[10000+500+500+...+500\] is the interest in \[{{15}^{th}}\] year. (\[500\] is \[14\] added times)

Therefore, the amount in \[{{15}^{th}}\] year 

\[=\]Rs.\[10000+14\times \]Rs.\[500\]

\[=\]Rs.\[10000+\]Rs.\[7000\]

\[=\]Rs.\[17000\]

Rs.\[10000+500+500+...+500\] is the amount after \[20\] years. (\[500\] is \[20\] added times)

Therefore, the amount after \[20\] years

\[=\]Rs.\[10000+20\times \]Rs.\[500\]

\[=\]Rs.\[10000+\]Rs.\[10000\]

\[=\]Rs.\[20000\]

The total amount after \[20\] years is Rs.\[20000\].

31. A manufacturer reckons that the value of a machine, which costs him Rs. \[15625\], will depreciate each year by \[20%\]. Find the estimated value at the end of \[5\] years.

Ans:

The cost of the machine is Rs.\[15625\].

Every year machine depreciates by \[20%\].

Therefore, \[80%\] of the original cost ,i.e., \[\frac{4}{5}\] of the original cost is its value after every year.

Therefore, the value at the end of \[5\] years 

\[=15626\times \frac{4}{5}\times \frac{4}{5}\times ...\times \frac{4}{5}\]

\[=5\times 1024\]

\[=5120\]

Therefore, Rs.\[5120\] is the value of the machine at the end of \[5\] years.

32. \[150\] workers were engaged to finish a job in a certain number of days. \[4\] workers dropped out on second day, \[4\] more workers dropped out on third day and so on. It took \[8\] more days to finish the work. Find the number of days in which the work was completed.

Ans:

Let the number of days in which \[150\] workers finish the work be \[x\].

According to the conditions given in the question,

\[150x=150+146+142+...\left( x+8 \right)\]terms

With first term \[a=146\], common difference \[d=-4\] and number of turns as \[\left( x+8 \right)\] , the series \[150+146+142+...\left( x+8 \right)\]terms is an A.P. 

\[\Rightarrow 150x=\frac{\left( x+8 \right)}{2}\left[ 2\left( 150 \right)+\left( x+8-1 \right)\left( -4 \right) \right]\]

\[\Rightarrow 150x=\left( x+8 \right)\left[ 150+\left( x+7 \right)\left( -2 \right) \right]\]

\[\Rightarrow 150x=\left( x+8 \right)\left( 150-2x-14 \right)\]

\[\Rightarrow 150x=\left( x+8 \right)\left( 136-2x \right)\]

\[\Rightarrow 75x=\left( x+8 \right)\left( 68-x \right)\]

\[\Rightarrow 75x=68x-{{x}^{2}}+544-8x\]

\[\Rightarrow {{x}^{2}}+75x-60x-544=0\]

\[\Rightarrow {{x}^{2}}+15x-544=0\]

\[\Rightarrow {{x}^{2}}+32x-17x-544=0\]

\[\Rightarrow x\left( x+32 \right)-17\left( x+32 \right)=0\]

\[\Rightarrow \left( x-17 \right)\left( x+32 \right)=0\]

\[\Rightarrow x=17\] or \[x=-32\]

We know that \[x\] cannot be negative.

So, \[x=17\].

Therefore, \[17\] is the number of days in which the work was completed. Then the required number of days \[=\left( 17+8 \right)=25\] .

Benefits of Referring to NCERT Solutions Provided by Vedantu

We at Vedantu provide you with all the study material that you are searching for in your preparation and examination. Here you will get NCERT Solutions, syllabus, previous year's paper’s solutions for exams, important questions etc.

NCERT Solutions are provided here as per the CBSE guidelines. Solutions of NCERT are prepared by highly-experienced teachers and subject-matter experts. The benefits of NCERT Solutions at Vedantu are as follows.

• The Solution in Easy Language

The NCERT Solutions of each chapter are prepared by different experts and scholars in the subject matter. The study materials offered by Vedantu are made available to students after rigorous research to ensure that all the given inputs are authentic and to the point.

• Focus on Fundamental Concepts

NCERT Class 11 chapter-wise solutions not only cover all the topics in the syllabus but also vividly describe all the fundamental and basic concepts required to understand these topics. 

• Sufficient Material to Practice

Preparation for any exam is incomplete without practice. Students are required to practice questions in order to perform well in examinations.

• Important Topics

Important topics given in the chapter are discussed from the point of view of examination. You can get questions along with their solutions on important topics of each chapter from our website.

• Better Preparation

Class 11 Maths NCERT Solutions will resolve the doubts of students quickly and their preparation for examinations will be boosted. With the help of these NCERT Solutions, students will grab complex concepts quickly.

NCERT Solutions for Class 11 Maths Chapters

• Chapter 1 - Sets

• Chapter 2 - Relations and Functions

• Chapter 3 - Trigonometric Functions

• Chapter 4 - Principle of Mathematical Induction

• Chapter 5 - Complex Numbers and Quadratic Equations

• Chapter 6 - Linear Inequalities

• Chapter 7 - Permutations and Combinations

• Chapter 8 - Binomial Theorem

• Chapter 9 - Sequences and Series

• Chapter 10 - Straight Lines

• Chapter 11 - Conic Sections

• Chapter 12 - Introduction to Three Dimensional Geometry

• Chapter 13 - Limits and Derivatives

• Chapter 14 - Mathematical Reasoning

• Chapter 15 - Statistics

• Chapter 16 - Probability

NCERT Solutions for Class 11 Maths Chapter 9 Sequences and Series PDF – Free Download

Most of the students primarily focus on getting full marks on Mathematics since there is a scope for that if one can master the core concepts. However, grasping the subject becomes beneficial in a lot of ways, especially for students who want to opt for a career related to this.

Ch 9 Maths Class 11 has lots of important sections regarding Sequence and Series that are required in further studies. Sequence and Series Class 11 Solutions help to explore these sections and implement better techniques to solve complex problems during examinations.

NCERT Solutions Class 11 Maths Chapter 9 PDF contains a set of unique questions and advanced solutions that help to finish the paper on time. The PDF version can be easily accessed from Vedantu app for free to help you find accurate solutions.

Sub-sections of Sequences and Series Class 11

Right kind of study materials can always make the learning experience smoother, and Sequence and Series NCERT Solutions can be considered as one of them. There are certain areas in this chapter that require extra attention and practice to grasp properly.

Some of these sections are:

• The first part mainly starts with an introduction to this entire chapter and elaborates on what you will get to learn along the way.

• Second part deals with Sequences which is a part of Algebra and explains how the arrangement of numbers works by following some relevant rules. There is a Finite and Infinite Sequence which is important to go through in order to understand this entire section.

• The next part focuses on Series, which is the sum of some set of numbers arranged in a specific order. Being another core part of Algebra, the series section contains interesting numerical problems which cannot be appropriately solved if you do not delve deeper.

• Then Arithmetic Progression comes in the picture that represents the difference of two consecutive terms, which is a constant.

• In the fifth section, students get to learn about Geometric Progression. It is more like Arithmetic Progression, but here it deals with the ratio of two consecutive terms which is also constant.

• Sixth section asserts the main differences between these two previous sections and related information. In order to solve relevant problems related to this chapter, you should grasp this section.

We cover all exercises in the chapter given below:

NCERT Solution Class 11 Maths of Chapter 9 Exercises

Topics Covered in the Chapter 9 Sequence and Series

Topics covered in the Chapter 9 Sequence and Series are as follows.

• Arithmetic Progression (A.P.)

• Arithmetic Mean (A.M.)

• Geometric Progression (G.P)

• The general term of a G.P

• Sum to n terms of a G.P

• Geometric Mean (G.M)

• Relationship Between A.M. and G.M.

• Sum to n Terms of Special Series, including,

• Sum of the first n natural numbers.

• Sum of squares of the first n natural numbers.

• Sum of cubes of the first n natural numbers.

NCERT Class 11 Maths Chapter 9 Weightage 2024-25

NCERT Class 11 Maths Chapter 9 Sequences and Series is a part of unit II Algebra. The unit consists of 6 chapters and is the second most important unit of the NCERT Syllabus for Class 11 maths. The complete unit constitutes a total of 13 marks out of 40 in Term 1 examinations of CBSE Maths Class 11, while it takes up 11 marks in Term 2.

Chapter 9 - Sequences and Series is included in the syllabus of the Class 11 Maths Term 1 exam. However, we do not recommend referring to any chapter-wise weightage as it keeps on changing and all the chapters are important.

Benefits of NCERT Solutions for Class 11 Maths Chapter 9

Since there are certain difficult areas in the chapter, students can face difficulties while solving exercise questions. However, it is advised to go through each section of this chapter and solve exercise questions alongside to find proper relevance. The major benefits include:

• Class 11 Maths NCERT Solutions Sequence and Series are crafted by academic professionals so that students do not struggle with finding accurate answers.

• Sequence and Series Class 11 NCERT solutions are presented in a concise structure so that students get the relevance once they are done with each section.

• The fast-solving method is the most important feature Sequence and Series Class 11 NCERT Solutions consist of.

• Moreover, Class 11 Maths NCERT Solutions Chapter 9 maintains an easy approach for better understanding and eliminating conceptual errors.

When it comes to learning better solution methods, Sequences and Series Class 11 NCERT Solutions contain specific features that benefit in the long run. So download it now

Conclusion 

The NCERT Solutions for Class 11 Maths Chapter 9 is important for students to improve their performance. The important topics covered in the chapter are given above and the solutions are prepared by our experts in easy language so that students can easily understand them. It will surely help students. If a student required any further study material, he/she can easily it download from our website.