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

Last updated date: 10th Aug 2024
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## NCERT Solutions for Maths Chapter 8 Class 11 Miscellaneous Exercise Solutions - Free PDF Download

NCERT Solutions for Class 11 Maths Chapter 8 Sequences And Series includes solutions to all Miscellaneous Exercise problems. The Miscellaneous Exercise combines key topics like arithmetic and geometric progressions, the sum of terms, and properties of sequences. This focuses on recognising patterns, and using formulas to find specific terms and methods to sum series.

Table of Content
1. NCERT Solutions for Maths Chapter 8 Class 11 Miscellaneous Exercise Solutions - Free PDF Download
2. Access NCERT Solutions Class 11 Maths Chapter 8 Sequences and Series
2.1Miscellaneous Exercise
3. Class 11 Maths Chapter 8: Exercises Breakdown
4. CBSE Class 11 Maths Chapter 8 Other Study Materials
5. Chapter-Specific NCERT Solutions for Class 11 Maths
FAQs

Class 11 Maths NCERT Solutions helps students understand how sequences grow and how to add up terms in a series. To perform well on the board exam, start practising from the start by downloading the FREE CBSE Class 11 Maths Syllabus.

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## Access NCERT Solutions Class 11 Maths Chapter 8 Sequences and Series

### 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)

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

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

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

 Exercise Number of Questions Exercise 8.1 14 Questions & Solutions Exercise 8.2 32 Questions & Solutions

## CBSE Class 11 Maths Chapter 8 Other Study Materials

 S. No Important Links for Chapter 8 Sequences and Series 1 Class 11 Sequences and Series Revision Notes 2 Class 11 Sequences and Series Important Questions 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 Miscellaneous Exercise Chapter 8 - Sequences and Series

1. What are the key topics in the NCERT Sequence And Series Class 11 Miscellaneous Exercise Solutions?

In NCERT Sequence And Series Class 11 Miscellaneous Exercise Solutions, the key topics include arithmetic progression (AP), geometric progression (GP), and their interrelationship. Understanding series sums and formulae is crucial. These form the foundation of the questions in this exercise. Mastery of these concepts is essential for solving complex problems and applying the correct methods to find sums and general terms.

2. What is the importance of solving NCERT Class 11 Maths Chapter 8 Miscellaneous Exercise Solutions?

NCERT Class 11 Maths Chapter 8 Miscellaneous Exercise Solutions offers a comprehensive mix of all the concepts covered in the chapter. It is vital for reinforcing understanding and applying various formulae and theorems in different contexts. This practice is critical for exam preparation as it hones problem-solving skills and ensures familiarity with different types of questions.

3. How should I approach solving these questions in NCERT Class 11 Maths Chapter 8 Miscellaneous Exercise Solutions?

Begin by reviewing the fundamental formulae for AP and GP. Practice deriving terms and sums for different series. Understand the proofs and logic behind these formulae, which will aid in tackling complex problems. Regular practice and reviewing different types of problems are key strategies for mastering this exercise.

4. What common mistakes should I avoid while solving Class 11 Maths Chapter 8 Miscellaneous Exercise Solutions?

Avoid memorizing formulae without understanding the underlying logic. Misinterpreting whether a series is an AP or GP is a common mistake, so read questions carefully. Regular practice is essential to avoid calculation errors and to improve problem-solving speed and accuracy.

5. Are there any specific strategies for exam preparation in the NCERT Miscellaneous Exercise in Chapter 8 Class 11?

In NCERT Miscellaneous Exercise On Chapter 8 Class 11, practice questions from previous years’ exams and sample papers focusing on this chapter. Time yourself to improve speed and accuracy. Utilize resources like Vedantu for detailed explanations and problem-solving techniques. Consistent practice and understanding are the best strategies for success.

6. What types of problems are included in the Miscellaneous Exercise On Chapter 8 Class 11?

The Miscellaneous Exercise On Chapter 8 Class 11 includes a variety of problems such as expanding binomial expressions, finding specific terms in expansions, and solving real-life application problems using the binomial theorem.

7. How can I determine which binomial expansion formula to use in the Miscellaneous Exercise On Chapter 8 Class 11?

To determine the appropriate binomial expansion formula, identify the form of the expression and the power to which it is raised.

8. What strategies can help me solve problems in the Class 11 Maths Sequence And Series Miscellaneous Exercise more effectively?

Understanding the properties of binomial coefficients, practising with a variety of problems, and reviewing solved examples can help you solve these problems more effectively.

9. How can I verify my answers for the Class 11 Maths Sequence And Series Miscellaneous Exercise problems?

You can verify your answers by cross-checking with the NCERT solutions, re-evaluating the problem step-by-step, or using online calculators designed for binomial expansions.

10. What is the significance of practising Class 11 Maths Sequence And Series Miscellaneous Exercise problems for exams?

Practicing these problems is crucial as they cover a broad range of concepts and problem types, helping you prepare for different question formats that may appear in exams.

11. Can the concepts from the Class 11 Maths Sequence And Series Miscellaneous Exercise be applied to competitive exams?

Yes, the concepts and problem-solving techniques learned from the Miscellaneous Exercise are very useful for competitive exams, where similar types of binomial theorem questions may appear.