Miscellaneous Exercise (Revised) - Chapter 9 - Sequences & Series - Ncert Solutions class 11 - Maths
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Chapter 8 - Sequences & Series | NCERT Solutions for Class 11 Maths
Miscellaneous Exercise Question 1.
If $f$ is a function satisfying $f(x+y)=f(x) f(y)$ for all $x, y \in \mathrm{N}$ such that $f(1)$ $=3$ and $\sum_{x=1}^n f(x)=120$, find the value of $n$.
Answer.
Given $f(1)=3$ and $f(x+y)=f(x) f(y)$ for all $x, y \in \mathrm{N}$ $\qquad$
Putting $x=1, y=1$ in eq. (i), $f(1+1)=f(1) f(1)$
$\Rightarrow f(2)=3 \times 3=9$
Putting $x=1, y=2$ in eq. (i), $f(1+2)=f(1) f(2)$
$
\Rightarrow f(3)=3 \times 9=27
$
Putting $x=1, y=3$ in eq. (i), $f(1+3)=f(1) f(3)$
$
\Rightarrow f(4)=3 \times 27=81
$
Now, $\sum_{x=1}^n f(x)=120$
$
\Rightarrow f(1)+f(2)+f(3)+\ldots \ldots . . f(n)=120
$
$
\Rightarrow 3+9+27+81+\ldots \ldots \text { up to } n \text { term }=120
$
$\begin{aligned}
& \Rightarrow \frac{3\left(3^n-1\right)}{3-1}=120 \\
& \Rightarrow 3\left(3^n-1\right)=240 \\
& \Rightarrow 3^n-1=80 \\
& \Rightarrow 3^n=81 \\
& \Rightarrow 3^n=3^4
\end{aligned}$
Miscellaneous Exercise Question 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.
Answer.
Given: $a=15, r=2$ and $S_n=315$
$
\begin{aligned}
& \therefore \mathrm{S}_n=\frac{a\left(r^n-1\right)}{\gamma-1} \\
& \Rightarrow 315=\frac{5\left(2^n-1\right)}{2-1} \\
& \Rightarrow \frac{315}{5}=2^n-1 \\
& \Rightarrow 2^n-1=63 \\
& \Rightarrow 2^n=64=2^6 \\
& \Rightarrow n=6 \\
& \therefore a_6=a r^{6-1}=5 \times 2^5=5 \times 32=160
\end{aligned}
$
Hence the number of terms $=6$ and the last term $=160$
Miscellaneous Exercise Question 3.
The first term of a G.P. is 1 . The sum of the third term and fifth term is 90 . Find common ratio of G.P.
Answer.
Given: $a=1$ and $a_3+a_5=90$
$
\begin{aligned}
& \Rightarrow a r^2+a r^4=90 \\
& \Rightarrow a\left(r^2+r^4\right)=90 \\
& \Rightarrow 1 \times\left(r^2+r^4\right)=90
\end{aligned}
$
$
\begin{aligned}
& \Rightarrow r^2+r^4=90 \\
& \Rightarrow r^4+r^2-90=0 \\
& \Rightarrow r^2=\frac{-1 \pm \sqrt{(1)^2-4 \times(-90) \times 1}}{2 \times 1} \\
& =\frac{-1 \pm \sqrt{1+360}}{2}=\frac{-1 \pm \sqrt{361}}{2} \\
& =\frac{-1 \pm 19}{2} \\
& \Rightarrow r^2=\frac{-1+19}{2}=\frac{18}{2}=9 \text { or } r^2=\frac{-1-19}{2}=\frac{-20}{2}=-10 \text { which is not possible }
\end{aligned}
$
Therefore, common ratio is $\gamma= \pm 3$
Miscellaneous Exercise Question 4.
The sum of 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.
Answer.
Let $a$ : $a r$, $a r^2$ be three numbers in G.P., therefore, $a+a r+a r^2=56$
$
\Rightarrow a\left(1+r+\mu^2\right)=56
$
According to question, $a-1, a r-7, a r^2-21$ are in A.P.
$
\begin{aligned}
& \therefore(a r-7)-(a-1)=\left(a r^2-21\right)-(a r-7) \\
& \Rightarrow a r-7-a+1=a r^2-21-a r+7 \\
& \Rightarrow a r-a-6=a r^2-a r-14 \\
& \Rightarrow a r^2-2 a r+a=8 \\
& \Rightarrow a\left(r^2-2 r+1\right)=8 \ldots . . \text { (ii) }
\end{aligned}
$
Dividing eq. (i) by eq. (ii), $\frac{a\left(1+r+r^2\right)}{a\left(r^2-2 r+1\right)}=\frac{56}{8}$
$
\begin{aligned}
& \Rightarrow 1+r+r^2=7 r^2-14 r+7 \\
& \Rightarrow 6 r^2-15 r+6=0 \\
& \Rightarrow 2 r^2-5 r+2=0 \\
& \Rightarrow r=\frac{-(-5) \pm \sqrt{(-5)^2-4 \times 2 \times 2}}{2 \times 2} \\
& =\frac{5 \pm \sqrt{25-16}}{4}=\frac{5 \pm \sqrt{9}}{4}=\frac{5 \pm 3}{4} \\
& \Rightarrow r=\frac{5+3}{4}=\frac{8}{4}=2 \text { or } r=\frac{5-3}{4}=\frac{2}{4}=\frac{1}{2}
\end{aligned}
$
Putting $r^{\prime}=2$ in eq. (i), $a\left(1+2+2^2\right)=56$
$
\Rightarrow a=\frac{56}{7}=8
$
Then the required numbers are $8,16,32$.
Putting $r=\frac{1}{2}$ in eq. (i), $a\left(1+\frac{1}{2}+\frac{1}{4}\right)=56$
$
\begin{aligned}
& \Rightarrow a \times \frac{7}{4}=56 \\
& \Rightarrow a=32
\end{aligned}
$
Then the required numbers are $32,16,8$.
Miscellaneous Exercise Question 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.
Answer.
Let the number of terms be $2 n$ then we have the number of odd terms is $n$
Let the G.P be $a, a r^*: a r^2$ $\qquad$ $a r^{2 n-1}$
Then the odd terms $a, a r^2, a r^4, a r^6$. $\qquad$ form a G.P
$
\therefore S_{2 n}=\frac{a\left(r^{2 n}-1\right)}{r-1} \text { and } S_n=a\left[\frac{\left(r^2\right)^n-1}{r^2-1}\right]
$
According to question, $S_{2 n}=5 S_n$
$
\begin{aligned}
& \Rightarrow a\left[\frac{r^{2 n}-1}{r-1}\right]=5 a\left[\frac{\left(r^2\right)^n-1}{r^2-1}\right] \\
& \Rightarrow \frac{1}{r-1}=\frac{5}{r^2-1} \\
& \Rightarrow r+1=5 \\
& \Rightarrow r=4
\end{aligned}
$
Miscellaneous Exercise Question 6.
If $\frac{a+b x}{a-b x}=\frac{b+c x}{b-c x}=\frac{c+d x}{c-d x}(x \neq 0)$, then show that $a, b, c$ and $d$ are in G.P.
Answer.
Taking $\frac{a+b x}{a-b x}=\frac{b+c x}{b-c x}$
$
\begin{aligned}
& \Rightarrow(a+b x)(b-c x)=(b+c x)(a-b x) \\
& \Rightarrow a b-a c x+b^2 x-b x^2=a b-b^2 x+a c x-b c x^2 \\
& \Rightarrow 2 b^2 x=2 a c x \\
& \Rightarrow b^2=a c \\
& \Rightarrow \frac{b}{a}=\frac{c}{b}
\end{aligned}
$
Taking $\frac{b+c x}{b-c x}=\frac{c+d x}{c-d x}$
$
\begin{aligned}
& \Rightarrow(b+c x)(c-d x)=(c+d x)(b-c x) \\
& \Rightarrow 2 c^2 x=2 b d x \\
& \Rightarrow c^2=b d \\
& \Rightarrow \frac{c}{b}=\frac{d}{c} \ldots \ldots \ldots . \text { (ii) }
\end{aligned}
$
From eq. (i) and (ii), $\frac{b}{a}=\frac{c}{b}=\frac{d}{c}$
Miscellaneous Exercise Question 7.
Let $S$ be the sum, $P$ the product and $R$ the sum of reciprocals of $n$ terms in a G.P. Prove that $\mathrm{P}^2 \mathrm{R}^n=\mathrm{S}^n$.
Answer.
Let the G.P be $a, a r, a r^2, a r^3 \ldots \ldots \ldots \ldots \ldots \ldots \ldots, a r^{n-1}$
Here $\mathrm{S}=\frac{a\left(r^n-1\right)}{r-1}$
$
\mathrm{P}=a \cdot a r^2 \cdot a r^2 \ldots \ldots \ldots a r^{n-1}=a^n \cdot r^{1+2+3+\ldots \ldots .+(n-1)}=a^n \cdot r^{\frac{n(n-1)}{2}}
$
and $\mathrm{R}=\frac{1}{a}+\frac{1}{a r^r}+\frac{1}{a r^2}+\ldots \ldots \ldots \frac{1}{a r^{n-1}}=\frac{r^{n-1}+r^{n-2}+r^{n-3}+\ldots \ldots \ldots .+1}{a r^{n-1}}$
$
=\frac{1\left(r^n-1\right)}{r-1} \cdot \frac{1}{a r^{n-1}}=\frac{r^n-1}{a r^{n-1}(r-1)}
$
Now $p^2 R^n=\frac{a^{2 n} \cdot r^{n(n-1)}\left(r^n-1\right)^n}{a^n r^{n(n-1)}(r-1)^n}=\frac{a^n\left(r^n-1\right)^n}{(r-1)^n}=a^n\left(\frac{r^n-1}{r-1}\right)^n=S^n$
Hence proved.
Miscellaneous Exercise Question 8.
If $a\left(\frac{1}{b}+\frac{1}{c}\right) b\left(\frac{1}{c}+\frac{1}{a}\right) c\left(\frac{1}{a}+\frac{1}{b}\right)$ are in A.P., prove that $a, b, c$ are in A.P.
Answer.
Given: $a\left(\frac{1}{b}+\frac{1}{c}\right), b\left(\frac{1}{c}+\frac{1}{a}\right), c\left(\frac{1}{a}+\frac{1}{b}\right)$ are in A.P.
$\Rightarrow a\left(\frac{b+c}{b c}\right), b\left(\frac{c+a}{c a}\right), c\left(\frac{a+b}{a b}\right)$ are in A.P.
$\Rightarrow \frac{a b+a c}{b c}=\frac{b c+a b}{c a}=\frac{a c+b c}{a b}$ are in A.P.
$\Rightarrow \frac{1}{b c}=\frac{1}{c a}=\frac{1}{a b} \text { are in A.P.[Dividing each fraction by } a b+b c+c a \text { ] }$
$\Rightarrow \frac{a b c}{b c}, \frac{a b c}{c a}, \frac{a b c}{a b}$ are in A.P.[Multiplying each fraction by $a b c$ ]
$\Rightarrow a, b, c$ are in A.P.
Miscellaneous Exercise Question 9.
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.
Answer.
Given: $a, b, c, d$ are in G.P.
To prove: $\left(a^n+b^n\right)=\left(b^n+c^n\right)=\left(c^n+d^n\right)$ are in G.P.
$\Rightarrow \frac{b^n+c^n}{a^n+b^n}=\frac{c^n+d^n}{b^n+c^n}$
Let $\frac{b}{a}=\frac{c}{b}=\frac{d}{c}=k$
$\therefore \frac{b}{a}=k$
$\Rightarrow b=a k$
And $\frac{c}{b}=k$
$\Rightarrow c=b k=(a k) k=a k^2$
Also $\frac{d}{c}=k$
$
\Rightarrow d=c k=\left(a k^2\right) k=a k^3
$
Now, $\frac{b^n+c^n}{a^n+b^n}=\frac{c^n+d^n}{b^n+c^n}$
$
\Rightarrow \frac{(a k)^n+\left(a k^2\right)^n}{a^n+(a k)^n}=\frac{\left(a k^2\right)^n+\left(a k^3\right)^n}{(a k)^n+\left(a k^2\right)^n}
$
$
\begin{aligned}
& \Rightarrow \frac{a^n k^n+a^n k^{2 n}}{a^n+a^n k^n}=\frac{a^n k^{2 n}+a^n k^{3 n}}{a^n k^n+a^n k^{2 n}} \\
& \Rightarrow \frac{a^n k^n\left(1+k^n\right)}{a^n\left(1+k^n\right)}=\frac{a^n k^{2 n}\left(1+k^n\right)}{a^n k^n\left(1+k^n\right)} \\
& \Rightarrow k^n=k^n
\end{aligned}
$
Therefore, $\left(a^n+b^n\right) \cdot\left(b^n+c^n\right) \cdot\left(c^n+d^n\right)$ are in G.P.
Miscellaneous Exercise Question 10.
If $a$ and $b$ are the roots $x^2-3 x+p=0$ and $c=d$ are roots of $x^2-12 x+q=0$. where $a, b, c, d$ form a G.P. Prove that $(q+p):(q-p)=17: 15$.
Answer.
Let $\frac{b}{a}=\frac{c}{b}=\frac{d}{c}=k$
$
\begin{aligned}
& \therefore \frac{b}{a}=k \\
& \Rightarrow b=a k
\end{aligned}
$
And $\frac{c}{b}=k$
$
\Rightarrow c=b k=(a k) k=a k^2
$
Also $\frac{d}{c}=k$
$
\Rightarrow d=c k=\left(a k^2\right) k=a k^3
$
$\because a$ and $b$ are the roots $x^2-3 x+p=0$
$
\therefore a+b=\frac{-(-3)}{1}=3
$
$
\begin{aligned}
& \Rightarrow a+a k=3 \\
& \Rightarrow a(1+k)=3
\end{aligned}
$
And $a b=\frac{p}{1}$
$
\begin{aligned}
& \Rightarrow a(a k)=p \\
& \Rightarrow a^2 k=p \ldots
\end{aligned}
$
Also $c$ : $d$ are roots of $x^2-12 x+q=0$
$
\begin{aligned}
& \therefore c+d=\frac{-(-12)}{1}=12 \\
& \Rightarrow a k^2+a k^3=12 \\
& \Rightarrow a k^2(1+k)=12 \ldots \ldots . . .
\end{aligned}
$
And $c d=\frac{q}{1}$
$
\Rightarrow a k^2\left(a k^3\right)=q
$
$
\Rightarrow a^2 k^5=q
$
Dividing eq. (iii) by eq. (i), $\frac{a k^2(1+k)}{a(1+k)}=\frac{12}{3}$
$
\begin{aligned}
& \Rightarrow k^2=4 \\
& \Rightarrow k= \pm 2
\end{aligned}
$
Now $\frac{q+p}{q-p}=\frac{a^2 k^5+a^2 k}{a^2 k^5-a^2 k}=\frac{a^2 k\left(k^4+1\right)}{a^2 k\left(k^4-1\right)}$
$
=\frac{( \pm 2)^4+1}{( \pm 2)^4-1}=\frac{16+1}{16-1}=\frac{17}{15}
$
Therefore, $(q+p):(q-p)=17: 15$
Miscellaneous Exercise Question 11.
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) \text {. }
$
Answer.
Given: $\frac{a+b}{2}: \sqrt{a b}=m: n$
$
\Rightarrow \frac{a+b}{2 \sqrt{a b}}=\frac{m}{n}
$
By componendo and dividendo,
$
\begin{aligned}
& \frac{a+b+2 \sqrt{a b}}{a+b-2 \sqrt{a b}}=\frac{m+n}{m-n} \\
& \Rightarrow \frac{(\sqrt{a}+\sqrt{b})^2}{(\sqrt{a}-\sqrt{b})^2}=\frac{m+n}{m-n}
\end{aligned}
$
$
\Rightarrow \frac{\sqrt{a}+\sqrt{b}}{\sqrt{a}-\sqrt{b}}=\frac{\sqrt{m+n}}{\sqrt{m-n}}
$
Again by componendo and dividendo,
$
\begin{aligned}
& \frac{\sqrt{a}+\sqrt{b}+\sqrt{a}-\sqrt{b}}{\sqrt{a}+\sqrt{b}-\sqrt{a}+\sqrt{b}}=\frac{\sqrt{m+n}+\sqrt{m-n}}{\sqrt{m+n}-\sqrt{m-n}} \\
& \Rightarrow \frac{2 \sqrt{a}}{2 \sqrt{b}}=\frac{\sqrt{m+n}+\sqrt{m-n}}{\sqrt{m+n}-\sqrt{m-n}}
\end{aligned}
$
$
\begin{aligned}
& \Rightarrow \frac{a}{b}=\frac{(\sqrt{m+n}+\sqrt{m-n})^2}{(\sqrt{m+n}-\sqrt{m-n})^2} \\
& \Rightarrow \frac{a}{b}=\frac{m+n+m-n+2 \sqrt{(m+n)(m-n)}}{m+n+m-n-2 \sqrt{(m+n)(m-n)}} \\
& \Rightarrow \frac{a}{b}=\frac{2 m+2 \sqrt{(m+n)(m-n)}}{2 m-2 \sqrt{(m+n)(m-n)}} \\
& \Rightarrow \frac{a}{b}=\frac{m+\sqrt{(m+n)(m-n)}}{m-\sqrt{(m+n)(m-n)}}
\end{aligned}
$
Therefore, $a: b=\left(m+\sqrt{m^2-n^2}\right):\left(m-\sqrt{m^2-n^2}\right)$
Miscellaneous Exercise Question 12.
Find the sum of the following series up to $n$ terms:
(i) $5+55+555+$ $\qquad$
(ii) $.6+.66+.666+$ $\qquad$
Ans. (i) $\mathrm{S}_n=5+55+555+$ $\qquad$ up to $n$ terms
$=5[1+11+111+$ $\qquad$ up to $n$ terms]
$=\frac{5}{9}[9+99+999+$ $\qquad$ up uo $n$ terms]
$
\begin{aligned}
& =\frac{5}{9}\left[(10-1)+\left(10^2-1\right)+\left(10^3-1\right)+\ldots \ldots . . \text { up to } n \text { terms }\right] \\
& =\frac{5}{9}\left[\frac{10\left(10^n-1\right)}{10-1}-n\right] \\
& =\frac{5}{9}\left[\frac{10}{9}\left(10^n-1\right)-n\right] \\
& =\frac{50}{81}\left(10^n-1\right)-\frac{5}{9} n
\end{aligned}
$
$
\text { (ii) } \mathrm{S}_n=.6+.66+.666+
$
$\qquad$ up to $n$ terms
$
=6[.1+.11+.111+
$
$\qquad$ up to $n$ terms]
$
=\frac{6}{9}[.9+.99+.999+
$
$\qquad$ up uo $n$ terms]
$
=\frac{6}{9}\left[\frac{9}{10}+\frac{99}{100}+\frac{999}{1000}+\right.
$
$\qquad$ up to $n$ terms $]$
$
=\frac{6}{9}\left[\left(1-\frac{1}{10}\right)+\left(1-\frac{1}{10^2}\right)+\left(1-\frac{1}{10^3}\right)\right.
$
$\qquad$ up to $n$ terms
$\begin{aligned}
& =\frac{2}{3}\left[n-\frac{\frac{1}{10}\left(1-\frac{1}{10^n}\right)}{1-\frac{1}{10}}\right] \\
& =\frac{2}{3}\left[n-\frac{1}{9}\left(1-\frac{1}{10^n}\right)\right]
\end{aligned}$
$
=\frac{2 n}{3}-\frac{2}{27}\left(1-\frac{1}{10^n}\right)
$
Miscellaneous Exercise Question 13.
Find the $20^{\text {th }}$ term of the series $2 \times 4+4 \times 6+6 \times 8+$ $\qquad$ + terms.
Answer.
Given: $2 \times 4+4 \times 6+6 \times 8+$ $\qquad$ $+n$ terms
$
\begin{aligned}
& \therefore a_n=\left(n^{\text {th }} \text { term of } 2,4,6, \ldots \ldots .\right)\left(n^{\text {nn }} \text { term of } 4,6,8, \ldots \ldots . .\right) \\
& \Rightarrow a_n=[2+(n-1) 2][4+(n-1) 2]=2 n(2 n+2) \\
& \therefore a_{20}=2 \times 20(2 \times 20+2)=40 \times 42=1680
\end{aligned}
$
Miscellaneous Exercise Question 14.
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 the tractor cost him?
Answer.
Total cost of the tractor $=$ Rs. 12000 , Cash paid $=$ Rs. 6000
Balance to be paid $=12000-6000=$ Rs. 6000
Annual installment $=$ Rs. 500
$\therefore$ Number of installment $=\frac{6000}{500}=12$
Interest of $1^{\text {st }}$ installment $=\frac{6000 \times 12 \times 1}{100}=$ Rs. 720
Amount of $1^{\text {st }}$ installment $=500+720=$ Rs. 1220
Interest of $2^{\text {nd }}$ installment $=\frac{5500 \times 12 \times 1}{100}=$ Rs. 660
Amount of $2^{\text {nd }}$ installment $=500+660=$ Rs. 1160
Interest of $3^{\text {rd }}$ installment $=\frac{5000 \times 12 \times 1}{100}=$ Rs. 600
Amount of $3^{\text {rd }}$ installment $=500+600=$ Rs. 1100
$\therefore$ Sequence of installments is $1220,1160,1100$, $\qquad$ which is in A.P
Here, $a=1220, d=1160-1220=-60$ and $n=12$
$
\begin{aligned}
& \therefore \mathrm{S}_n=\frac{n}{2}[2 a+(n-1) d] \\
& =\frac{12}{2}[2 \times 1220+(12-1) \times(-60)] \\
& =6[2440-660]=\text { Rs. } 10680
\end{aligned}
$
Therefore, the total cost of tractor is $(10680+6000)=$ Rs. 16680 .
Miscellaneous Exercise Question 15.
Shams had 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?
Answer.
Total cost of the scooter $=$ Rs. 22000 , Cash paid $=$ Rs. 4000
Balance to be paid $=22000-4000=$ Rs. 18000
Annual installment $=$ Rs. 1000
$
\begin{aligned}
& \therefore \text { Number of installment }=\frac{18000}{1000}=18 \\
& \text { Interest of } 1^{\text {st }} \text { installment }=\frac{18000 \times 10 \times 1}{100}=\text { Rs. } 1800
\end{aligned}
$
Amount of $1^{\text {st }}$ installment $=1000+1800=$ Rs. 2800
Interest of $2^{\text {nd }}$ installment $=\frac{17000 \times 10 \times 1}{100}=$ Rs. 1700
Amount of $2^{\text {nd }}$ installment $=1000+1700=$ Rs. 2700
Interest of $3^{\text {rd }}$ installment $=\frac{16000 \times 10 \times 1}{100}=$ Rs. 1600
Amount of $3^{\text {rd }}$ installment $=1000+1600=$ Rs. 2600
$\therefore$ Sequence of installments is $2800,2700,2600$, $\qquad$ in A.P Here, $a=2800, d=2700-2800=-100$ and $n=18$
$
\begin{aligned}
& \therefore \mathrm{S}_n=\frac{n}{2}[2 a+(n-1) d]=\frac{18}{2}[2 \times 2800+(18-1) \times(-100)] \\
& =9[5600-1700]=\text { Rs. } 35100
\end{aligned}
$
Therefore, the total cost of tractor is $(35100+4000)=$ Rs. 39100 .
Miscellaneous Exercise Question 16.
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 $8^{\text {th }}$ set of letter is mailed.
Answer.
Total letters in the first set $=4$, Total letters in the second set $=4^2=16$
Total letters in the third set $=4^3=64$
$\therefore$ Sequence of letters is $4,16,64$, $\qquad$ in G.P.
Here $a=4, r=\frac{16}{4}=4$ and $n=8$
$
\begin{aligned}
& \therefore \mathrm{S}_n=\frac{a\left(r^\pi-1\right)}{r-1} \\
& =\frac{4\left(4^8-1\right)}{8-1} \\
& =\frac{4}{3}(65536-1) \\
& =\frac{4}{3} \times 65535=87380
\end{aligned}
$
Hence,total number of letters mailed $=87380$
The amount of postage on each letter $=50$ paise
Therefore total amount spent on postage $=87380 \times 0.50=$ Rs. 43690 .
Miscellaneous Exercise Question 17.
A man deposited Rs. 10000 in a bank at the rate of $5 \%$ simple interest annually. Find the amount in $15^{\text {th }}$ year since he deposited the amount and also calculate the total amount after 20 years.
Answer.
Total amount deposited $=$ Rs. 10000 , Rate of interest $=5 \%$ per annum
Interest of first year $=\frac{10000 \times 5 \times 1}{100}=$ Rs. 500
Here $a=10000, d=500$
$\therefore$ Amount in $15^{\text {th }}$ year $=a_{15}=10000+(15-1) \times 500=10000+7000=$ Rs. 17000
Total amount after 20 years $=$ Amount in the $21^{\text {st }}$ year $=a_{21}=10000+(21-1) 500$
$
=10000+10000=\text { Rs. } 20000
$
Miscellaneous Exercise Question 18.
A manufacturer reckons that the value of a machine, which cost him Rs. 15625 will depreciate each year by $20 \%$. Find the estimated value at the end of 5 years.
Answer.
Present value of the machine = Rs. 15625
Rate of depreciation $=20 \%$
After 1 year value of machine $=15625-15625 \times \frac{20}{100}=15625-3125=$ Rs. 12500
After 2 year value of machine $=12500-12500 \times \frac{20}{100}=12500-2500=$ Rs. 10000
After 3 year value of machine $=10000-10000 \times \frac{20}{100}=10000-2000=$ Rs. 8000
$\therefore$ Sequence of values of machine after depreciation is $12500,10000,8000, \ldots . .$. is a G.P.
Here $a=12500, r=\frac{10000}{12500}=\frac{4}{5}$
$
\therefore a_5=a r^4=12500 \times\left(\frac{4}{5}\right)^4=12500 \times \frac{256}{625}=\text { Rs. } 5120
$
Therefore, the value of machine at the end of 5 years is Rs. 5120 .
Miscellaneous Exercise Question 19.
150 workers were engaged to finish a job in a certain number of boys. 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.
Answer.
Number of workers on the first day $=150$
Number of workers on the second day $=150-4=146$
Number of workers on the third day $=146-4=142$
$\therefore$ Sequence of number of workers is 150, 146, 142, $\qquad$ in A.P.
Here $a=150, d=146-150=-4$
$\therefore$ Total number of workers required to finish the work in $n$ days
$
\begin{aligned}
& =\frac{n}{2}[2 \times 150+(n-1)(-4)] \\
& =\frac{n}{2}(300-4 n+4) \\
& =n(152-2 n) \ldots \ldots . .(\mathrm{i})
\end{aligned}
$
If no worker had dropped out, then the work should have finished in $(n-8)$ days with 150 workers on each day.
$\therefore$ Total number of workers required to finish the work in $(n-8)$ days $=150(n-8)$
From eq. (i) and (ii), $n(152-2 n)=150(n-8)$
$
\begin{aligned}
& \Rightarrow 152 n-2 n^2=150 n-1200 \\
& \Rightarrow 2 n^2-2 n-1200=0 \\
& \Rightarrow n^2-n-600=0
\end{aligned}
$
$
\Rightarrow(n-25)(n+24)=0
$
$\Rightarrow n=25$ and $n=-24$ which is not possible
Therefore, the work was completed in 25 days.
