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# NCERT Solutions for Class 11 Maths Chapter 8 - Sequences and Series

Last updated date: 13th Sep 2024
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## NCERT Chapter 8 Maths Sequences and Series Class 11 Solutions - Free PDF Download

The NCERT Maths Chapter 8 Sequence and Series Class 11 Solutions is all about understanding the order and pattern of numbers. The free PDF of Chapter 8 Class 11 Sequence and Series Maths Solutions is available on Vedantu, providing students with a better understanding of the problems. It covers solutions to every exercise in this chapter and is updated according to the latest CBSE syllabus. In this chapter, you will learn about different types of sequences and geometric sequences. Vedantu’s solutions provide step-by-step explanations to help you grasp these concepts easily. The clear and detailed solutions ensure that you understand how to approach and solve problems related to sequences and series.

Table of Content
1. NCERT Chapter 8 Maths Sequences and Series Class 11 Solutions - Free PDF Download
2. Glance on Class 11th Maths Chapter 8 - Sequences and Series
3. Access Exercise wise NCERT Solutions for Chapter 8 Maths Class 11
4. Exercises Under NCERT Solutions for Class 11 Maths Chapter 8 – Sequences and Series
5. Access NCERT Solutions for Class 11 Maths Chapter 8 – Sequences and Series
5.1Exercise 8.1
5.2Exercise 8.2
5.3Miscellaneous Exercise
6. Also you can Find the Solutions of all the Maths Chapters Below.
7. Overview of Deleted Syllabus for CBSE Class 11 Maths Chapter - Sequences and Series
8. Class 11 Maths Chapter 8: Exercises Breakdown
9. Conclusion
10. Other Study Material for CBSE Class 11 Maths Chapter 8
11. Chapter-Specific NCERT Solutions for Class 11 Maths
FAQs

## Glance on Class 11th Maths Chapter 8 - Sequences and Series

• Sequence, we mean an arrangement of numbers in definite order according to some rule. Also, we define a sequence as a function whose domain is the set of natural numbers or some subsets of the type $\{1,2,3, \ldots i k$.

• A sequence containing a finite number of terms is called a finite sequence A sequence is called infinite if it is not a finite sequence

• Let $a_1 a_2, a_3, \ldots$ be the sequence, then the sum expressed as $a_1+a_2+a_3+\ldots$. is called a series. A series is called finite series ifit has got finite number of terms

• There are two exercises (46 fully solved questions) and one Miscellaneous exercise(18 fully solved questions) in class 11th Maths Chapter 8 Sequences and Series.

## Access Exercise wise NCERT Solutions for Chapter 8 Maths Class 11

 Current Syllabus Exercises of Class 11 Maths Chapter 8 NCERT Solutions of Class 11 Maths Sequences and Series Exercise 8.1 NCERT Solutions of Class 11 Maths Sequences and Series Exercise 8.2 NCERT Solutions of Class 11 Maths Sequences and Series Miscellaneous Exercise
Competitive Exams after 12th Science

## Exercises Under NCERT Solutions for Class 11 Maths Chapter 8 – Sequences and Series

• Exercise 8.1: This exercise contains 14 fully solved questions. 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 8.2: This exercise contains 32 fully solved questions. 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.

• Miscellaneous Exercise: This exercise contains 18 fully solved questions. 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 8 – Sequences and Series

### Exercise 8.1

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 8.2

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.

1. $\sqrt{3},3,3\sqrt{3}...$ 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.

1. $\frac{1}{3},\frac{1}{9},\frac{1}{27},...$ 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)

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

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

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.

### Miscellaneous Exercise

1. 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 $n$.

2. 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 $6$.

3. 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.

4. 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.

5. 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.

6: 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.

7. 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, ${{P}^{2}}{{R}^{n}}={{S}^{n}}$.

8. 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.

9. 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.

10. 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)

$2a=\frac{2\sqrt{ab}}{n}\left( m+\sqrt{{{m}^{2}}-{{n}^{2}}} \right)$

$\Rightarrow a=\frac{\sqrt{ab}}{n}\left( m+\sqrt{{{m}^{2}}-{{n}^{2}}} \right)$

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 $a:b=\left( m+\sqrt{{{m}^{2}}-{{n}^{2}}} \right):\left( m-\sqrt{{{m}^{2}}-{{n}^{2}}} \right)$.

11. 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 $\frac{50}{81}\left( {{10}^{n}}-1 \right)-\frac{5n}{9}$ .

1. $.6+.66+.666.+...$

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 $\frac{2}{3}n-\frac{2}{27}\left( 1-{{10}^{-n}} \right)$ .

12. 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.

13. 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.$16680$.

14. 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.$39100$ .

15. 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.

16. 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.$20000$.

17. 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.

18. $150$ 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 $=\left( 17+8 \right)=25$ .

## Also you can Find the Solutions of all the Maths Chapters Below.

### 8.1 Introduction:

The word “sequence” is used in much the same way as it is in ordinary English. When we say that a collection of objects is listed in a sequence, we usually mean that the collection is ordered in such a way that it has an identified first member, second member, third member and so on.

### 8.2 Sequences:

A sequence is a function whose domain is the set of natural numbers. In simpler terms, it is an arrangement of numbers in a particular order, following a specific rule or pattern.

Types of Sequences:

There are two types of Sequences

1. Finite Sequence and

2. Infinite Sequence

Finite Sequence: A finite sequence is a sequence that has a definite number of terms. The number of terms in a finite sequence is countable and fixed. Once you reach the last term, the sequence ends.

For example:

Consider the sequence 3,6,9,12,15,18. This sequence has exactly 6 terms, so it is a finite sequence.

Infinite sequence: An infinite sequence is a sequence that continues indefinitely without terminating. It has an infinite number of terms and does not have a last term.

For example:

Consider the sequence 1, 2, 3, 4, ..., n. This sequence continues without end, so it is an infinite sequence.

### 8.3 Series:

A series is defined as the sum of the terms of a sequence. If you have a sequence of numbers, then the corresponding series is the result of adding these numbers together.

### 8.4 Geometric Progression(G.P.):

Geometric Progression (G.P.) is a sequence of numbers where each term after the first is obtained by multiplying the previous term by a fixed, non-zero number called the common ratio.

General form of G.P. is a, ar, ar^2, ar^3,.....

8.4.1 Arithmetic Mean: The Arithmetic Mean (A.M.) of a set of numbers is the sum of the numbers divided by the count of the numbers. It is commonly known as the average.

We can calculate A.M. by using the formula

A.M. = (a1 + a2 + a3 + ... + an) / n

8.4.2 Geometric Mean: The Geometric Mean (G.M.) of a set of numbers is the nth root of the product of the numbers, where n is the count of the numbers.

We can calculate G.M. by using the formula

G.M. = (a1 * a2 * a3 * ... * an)^(1/n)

### 8.5 Relationship between A.M. and G.M.:

Inequality: The geometric mean of a set of positive numbers is always less than or equal to the arithmetic mean of the same set of numbers.

G.M. ≤ A.M.

Equality Condition: The equality holds if and only if all the numbers in the set are equal.

## Overview of Deleted Syllabus for CBSE Class 11 Maths Chapter - Sequences and Series

 Chapter Dropped Topics Sequences and Series 8.4 Arithmetic Progression (A.P.) 8.7 Sum to n terms of special series Examples 21, 22 and 24 Question numbers - 1,2,3,4,5,6,12,15,16,20,23,24,25,26 from Miscellaneous Exercise In summary points 3 and 4

## Class 11 Maths Chapter 8: Exercises Breakdown

 Exercises Number of Questions Exercise 8.1 14 Questions and Solutions Exercise 8.2 32 Questions and Solutions Miscellaneous Exercise 18 Questions and Solutions

## Conclusion

NCERT Solutions for Class 11 Maths Chapter 8 Solutions Sequences and Series by Vedantu provides a comprehensive guide to understanding different types of sequences such as Geometric Progression (G.P.), along with their corresponding series. Key concepts include finding and understanding the relationship between Arithmetic Mean (A.M.) and Geometric Mean (G.M.). This chapter is crucial as it lays the foundation for higher mathematical concepts and applications in various fields. According to previous exam papers, around 5–6 questions are typically asked from this chapter, emphasizing the importance of mastering these topics for scoring well in exams. Detailed solutions and step-by-step explanations provided by Vedantu help students grasp these concepts effectively, preparing them thoroughly for their exams.

## Other Study Material for CBSE Class 11 Maths Chapter 8

 S. No Important Links for Chapter 8 Sequences and Series 1 Class 11 Sequences and Series Important Questions 2 Class 11 Sequences and Series Revision Notes 3 Class 11 Sequences and Series NCERT Exemplar Solution

## Chapter-Specific NCERT Solutions for Class 11 Maths

Given below are the chapter-wise NCERT Solutions for Class 11 Maths. Go through these chapter-wise solutions to be thoroughly familiar with the concepts.

## FAQs on NCERT Solutions for Class 11 Maths Chapter 8 - Sequences and Series

1. What Does Arithmetic Progression Consist of?

Progression is all about a sequence where the terms maintain a specific pattern. In the case of Arithmetic progression, there are two consecutive terms and their difference is constant.

In order to understand the concept better, students should go through Class 11 Maths 8th Chapter and particularly this section. It is relatively easier if one can explore the underlying areas of it. Even though the textbook contains sufficient information regarding the same, you can also opt for study materials to understand the functionalities better.

2. How Many Sections are There in Class 11 Maths Chapter 8 Solutions?

Chapter 8 of Class 11 Maths mainly consists of six important sections apart from the introductory part where the entire chapter has been introduced. It further explains the concepts regarding Sequence, Series, Arithmetic and Geometric Progression, correlation between them and sum related to n terms of special series.

There are four exercises as well to help you with advanced questions to make your exam preparation better. These sections essentially give away small details that are required to solve relevant problems.

3. How to get Full Marks in Sequence and Series Chapter?

The Sequence and Series chapter demands a lot of practice to get rid of conceptual errors since it contains lots of difficult sections. The underlying concepts can be a little tricky in the beginning; however, with the right kind of guidance and advanced techniques, achieving a good score is not that difficult.

If your goal is to attain full marks in every question related to this chapter, you should consider practising different kinds of tricky questions as well. Class 11 Maths NCERT solutions Sequence and Series contain various fast-solving techniques that keep you aware of the type of questions you will be facing in exams.

4. Why are NCERT Solutions Class 11 Maths Chapter 8 Important?

With the use of real-world examples, NCERT Solutions Class 11 Maths Chapter 8 teaches students how to grasp sequences in the most basic way possible. The content in NCERT books is compiled by specialists and is also approved by the CBSE board, making them extremely important. Furthermore, there is a good chance that questions from these NCERT Solutions will appear in the tests. See all the solutions on the Vedantu official website to solve the questions in Chapter 8.

5. Do I Need to Practice all Questions Provided in NCERT Solutions Class 11 Maths Sequences and Series?

For finding the sum of sequences as well as the nth term, the NCERT Solutions Class 11 Maths Sequences and Series requires quite a few formulas. To memorize them, students should practice all of the solved examples, as well as the practice questions to reinforce their understanding of the various types of progressions and, as a result, boost their computation speed. Practicing all the questions from Vedantu will help you in various ways.

6. How Many Questions are there in NCERT Solutions Class 11 Maths Chapter 8 Sequences and Series?

There are 74 problems in NCERT Solutions Class 11 Maths Chapter 8 Sequences and Series and are divided into three categories: easy, fairly easy, and extended format. They are organized into four exercises, each of which focuses on a different aspect of this chapter. These include a wide range of questions relating to A.P. and G.P., allowing students to go deeper into the subject. Chapter 8 is not tough to practice so you can visit Vedantu and start preparing the material.

7. What are the Important Topics Covered in NCERT Solutions Class 11 Maths Chapter 8?

The notion of sequences, arithmetic progression, geometric progression, Fibonacci series, the sum of specific natural number sequences involving squares and cube roots are all explained in NCERT Solutions Class 11 Maths Chapter 8. The issue of arithmetic and geometric mean has also been presented with a logical explanation. All the important topics are crucial from the exam point of view. These important topics need more focus while preparing for the exam. All the preparation material is available on the Vedantu website and the Vedantu app free of cost.

8. How CBSE Students can utilize NCERT Solutions Class 11 Maths Chapter 8 effectively?

After giving the chapter a thorough read, students should make it a habit to practice the solved problems every day. After then, they should begin answering the exercise questions, one area at a time, to obtain sufficient practice and confidence in all of the ideas. Students should make a note of the formulas involved in the topic of arithmetic and geometric progression, or they can refer to the highlight section at the end of the chapter. Students can effectively use the NCERT Solutions Class 11 Maths Chapter 8 in this manner.