NCERT Solutions for Class 11 Maths Chapter 8 Sequences and Series Miscellaneous Exercise

Miscellaneous Exercise

1. If f 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 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 n 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 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)

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

11. Find the sum of the following series up to \[n\] terms:

(i) \[5+55+555+...\]

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

(ii) \[.6+.66+.666.+...\]

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 instalments of Rs.\[500\] plus \[12%\] interest on the unpaid amount. How much will 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 a 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 an annual instalment 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 the 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 it to four different persons with the 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 letters 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 of mailing 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 the second day, \[4\] more workers dropped out on the 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 \[=\left( 17+8 \right)=25\] .

Conclusion

The NCERT Miscellaneous Exercise for Chapter 8, Sequences and Series, is important for understanding arithmetic and geometric progressions. It includes a variety of problems that help you practice and apply these concepts. Working through these exercises will improve your problem-solving skills and prepare you for exams. Regular practice will make students more confident in dealing with sequences and series. Focus on solving these problems to strengthen your understanding of the chapter and perform better in your studies.

Class 11 Maths Chapter 8: Exercises Breakdown

CBSE Class 11 Maths Chapter 8 Other Study Materials

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.

Important Related Links for CBSE Class 11 Maths