Chapter 8 Sequences And Series Miscellaneous Exercise
Miscellaneous Exercise On Chapter 8
1. If $f$ is a function satisfying $f(x+y)=f(x) f(y)$ for all $x, y \in \mathbf{N}$ such that $ f(1)=3 \text{ and } \sum _{x=1}^{n} f(x)=120 \text{, find the value of } n \text{. } $
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Answer :
It is given that,
$f(x+y)=f(x) \times f(y)$ for all $x, y \in N$.
$f(1)=3$
Taking $x=y=1$ in (1), we obtain
$f(1+1)=f(2)=f(1) f(1)=3 \times 3=9$
Similarly,
$f(1+1+1)=f(3)=f(1+2)=f(1) f(2)=3 \times 9=27$
$f(4)=f(1+3)=f(1) f(3)=3 \times 27=81$
$\therefore f(1), f(2), f(3), \ldots$, that is $3,9,27, \ldots$, forms a G.P. with both the first term and common ratio equal to 3 .
It is known that,
$ S_n=\dfrac{a(r^{n}-1)}{r-1} $
It is given that, $\sum _{x=1}^{n} f(x)=120$ $\therefore 120=\dfrac{3(3^{n}-1)}{3-1}$
$\Rightarrow 120=\dfrac{3}{2}(3^{n}-1)$
$\Rightarrow 3^{n}-1=80$
$\Rightarrow 3^{n}=81=3^{4}$
$\therefore n=4$
Thus, the value of $n$ is 4 .
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.
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Answer :
Let the sum of $n$ terms of the G.P. be 315 .
It is known that, $S_n=\dfrac{a(r^{n}-1)}{r-1}$
It is given that the first term $a$ is 5 and common ratio $r$ is 2 .
$\therefore 315=\dfrac{5(2^{n}-1)}{2-1}$
$\Rightarrow 2^{n}-1=63$
$\Rightarrow 2^{n}=64=(2)^{6}$
$\Rightarrow n=6$
$\therefore$ Last term of the G.P $=6^{\text{th }}$ term $=a r^{6 - 1}=(5)(2)^{5}=(5)(32)=160$
Thus, the last term of the G.P. is 160.
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.
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Answer :
Let $a$ and $r$ be the first term and the common ratio of the G.P. respectively.
$\therefore a=1$
$a_3=a r^{2}=r^{2}$ $a_5=a r^{4}=r^{4}$
$\therefore r^{2}+r^{4}=90$
$\Rightarrow r^{4}+r^{2}- 90=0$
$\Rightarrow r^{2}=\dfrac{-1+\sqrt{1+360}}{2}=\dfrac{-1 \pm \sqrt{361}}{2}=\dfrac{-1 \pm 19}{2}=-10$ or 9
$\therefore r= \pm 3$
(Taking real roots)
Thus, the common ratio of the G.P. is $\pm 3$.
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.
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Answer :
Let the three numbers in G.P. be $a$, $a r$, and $a r^{2}$.
From the given condition, $a+a r+a r^{2}=56$
$\Rightarrow a(1+r+r^{2})=56$
$\Rightarrow a=\dfrac{56}{1+r+r^{2}}$
a - 1 , ar - $7, a r^{2} - 21$ forms an A.P.
$ \therefore (ar - 7)-(a-1) = (ar^2 - 21) - (ar - 7) $
$\Rightarrow ar$ - a - $6=a r^{2} -$ ar - 14
$\Rightarrow a r^{2} - 2 a r+a=8$
$\Rightarrow a r^{2}- a r- a r+a=8$
$\Rightarrow a(r^{2}+1 - 2 r)=8$
$\Rightarrow a(r- 1)^{2}=8$.
$\Rightarrow \dfrac{56}{1+r+r^{2}}(r-1)^{2}=8$
[Using (1)]
$\Rightarrow 7(r^{2} - 2 r+1)=1+r+r^{2}$
$\Rightarrow 7 r^{2} - 14 r+7$ - 1 - $r$ - $r^{2}=0$
$\Rightarrow 6 r^{2} - 15 r+6=0$
$\Rightarrow 6 r^{2} - 12 r - 3 r+6=0$
$ \therefore (6ar - 7)-(a-1) = (ar^2 - 21) - (ar - 7) $
$\Rightarrow(6 r- 3)(r$- 2$)=0$ $\therefore r=2, \dfrac{1}{2}$
When $r=2, a=8$
When $r=\dfrac{1}{2}, a=32$
Therefore, when $r=2$, the three numbers in G.P. are 8, 16, and 32.
When $r=\dfrac{1}{2}$, the three numbers in G.P. are 32, 16, and 8.
Thus, in either case, the three required numbers are 8,16 , and 32 .
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.
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Answer :
Let the G.P. be $T_1, T_2, T_3, T_4, \ldots T _{2 n}$.
Number of terms $=2 n$
According to the given condition,
$T_1+T_2+T_3+\ldots+T _{2 n}=5[T_1+T_3+\ldots+T _{2_n{-1}}]$
$\Rightarrow T_1+T_2+T_3+ \ldots +T_{2 n}- 5[T_1+T_3+\ldots + T_{2n}] =0$
$\Rightarrow T_2+T_4+\ldots+T _{2 n}=4[T_1+T_3+\ldots+T _{2 n {-1}}]$
Let the G.P. be $a, a r, a r^{2}, a r^{3}, \ldots$
$\therefore \dfrac{ar(r^{n}-1)}{r-1}=\dfrac{4 \times a(r^{n}-1)}{r-1}$
$\Rightarrow a r=4 a$
$\Rightarrow r=4$
Thus, the common ratio of the G.P. is 4.
6. If $\dfrac{a+b x}{a-b x}=\dfrac{b+c x}{b-c x}=\dfrac{c+d x}{c-d x}(x \neq 0)$, then show that $a, b, c$ and $d$ are in G.P.
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Answer:
It is given that,
$$ \begin{align*} & \dfrac{a+b x}{a-b x}=\dfrac{b+c x}{b-c x} \\ & \Rightarrow(a+b x)(b-c x)=(b+c x)(a-b x) \\ & \Rightarrow a b-a c x+b^{2} x-b c x^{2}=a b-b^{2} x+a c x-b c x^{2} \\ & \Rightarrow 2 b^{2} x=2 a c x \\ & \Rightarrow b^{2}=a c \\ & \Rightarrow \dfrac{b}{a}=\dfrac{c}{b} \tag{1} \end{align*} $$
Also, $\dfrac{b+c x}{b-c x}=\dfrac{c+d x}{c-d x}$
$\Rightarrow(b+c x)(c-d x)=(b-c x)(c+d x)$
$\Rightarrow b c-b d x+c^{2} x-c d x^{2}=b c+b d x-c^{2} x-c d x^{2}$
$\Rightarrow 2 c^{2} x=2 b d x$
$\Rightarrow c^{2}=b d$
$\Rightarrow \dfrac{c}{d}=\dfrac{d}{c}$
From (1) and (2), we obtain
$ \dfrac{b}{a}=\dfrac{c}{b}=\dfrac{d}{c} $
Thus, $a, b, c$, 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 $P^{2} R^{n}=S^{n}$.
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Answer :
Let the G.P. be $a, a r, a r^{2}, a r^{3}, \ldots a r^{n-1} \ldots$
According to the given information,
$ \begin{aligned} & S=\dfrac{a(r^{n}-1)}{r-1} \\ & P=a^{n} \times r^{1+2+\ldots+n-1} \\ & =a^{n} r^{\dfrac{n(n-1)}{2}} \\ & {[\because \text{ Sum of first } n \text{ natural numbers is } n \dfrac{(n+1)}{2}]} \\ & R=\dfrac{1}{a}+\dfrac{1}{a r}+\ldots+\dfrac{1}{a r^{n-1}} \\ & =\dfrac{r^{n-1}+r^{n-2}+\ldots r+1}{a r^{n-1}} \\ & =\dfrac{1(r^{n}-1)}{(r-1)} \times \dfrac{1}{a r^{n-1}} \quad[\because 1, r, \ldots r^{n-1} \text{ forms a G.P }] \\ & =\dfrac{r^{n}-1}{a r^{n-1}(r-1)} \\ & \therefore P^{2} R^{n}=a^{2 n} r^{n(n-1)} \dfrac{(r^{n}-1)^{n}}{a^{n} r^{n(n-1)}(r-1)^{n}} \\ & =\dfrac{a^{n}(r^{n}-1)^{n}}{(r-1)^{n}} \\ & =[\dfrac{a(r^{n}-1)}{(r-1)}]^{n} \\ & =S^{n} \end{aligned} $
Hence, $P^{2} R^{n}=S^{n}$
8. If $a, b, c, d$ are in G.P, prove that $(a^{n}+b^{n}),(b^{n}+c^{n}),(c^{n}+d^{n})$ are in G.P.
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Answer :
It is given that $a, b, c$, and $d$ are in G.P.
$\therefore b^{2}=a c$…
$c^{2}=b d$
$a d=b c$
It has to be proved that $(a^{n}+b^{n}),(b^{n}+c^{n}),(c^{n}+d^{n})$ are in G.P. i.e.,
$(b^{n}+c^{n})^{2}=(a^{n}+b^{n})(c^{n}+d^{n})$
Consider L.H.S.
$(b^{n}+c^{n})^{2}=b^{2 n}+2 b^{n} c^{n}+c^{2 n}$
$=(b^{2})^{n}+2 b^{n} c^{n}+(c^{2})^{n}$
$=(a c)^{n}+2 b^{n} c^{n}+(b d)^{n}[$ Using (1) and (2)]
$=a^{n} c^{n}+b^{n} c^{n}+b^{n} c^{n}+b^{n} d^{n}$
$=a^{n} c^{n}+b^{n} c^{n}+a^{n} d^{n}+b^{n} d^{n}$ [Using (3)]
$=c^{n}(a^{n}+b^{n})+d^{n}(a^{n}+b^{n})$
$=(a^{n}+b^{n})(c^{n}+d^{n})$
$=$ R.H.S.
$\therefore(b^{n}+c^{n})^{2}=(a^{n}+b^{n})(c^{n}+d^{n})$
Thus, $(a^{n}+b^{n}),(b^{n}+c^{n})$, and $(c^{n}+d^{n})$ are in G.P.
9. If $a$ and $b$ are the roots of $x^{2}-3 x+p=0$ and $c, d$ are roots of $x^{2}-12 x+q=0$, where $a, b, c, d$ form a G.P. Prove that $(q+p):(q-p)=17: 15$.
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Answer :
It is given that $a$ and $b$ are the roots of $x^2-3 x+p=0$
$\therefore a+b=3$ and $a b=p$
Also, $c$ and $d$ are the roots of $x^{2}-12 x+q=0$
$\therefore c+d=12$ and $c d=q$..
It is given that $a, b, c, d$ are in G.P.
Let $a=x, b=x r, c=x r^{2}, d=x r^{3}$
From (1) and (2), we obtain
$x+x r=3$
$\Rightarrow x(1+r)=3$
$x r^{2}+x r^{3}=12$
$\Rightarrow x r^{2}(1+r)=12$
On dividing, we obtain
$\dfrac{x r^{2}(1+r)}{x(1+r)}=\dfrac{12}{3}$
$\Rightarrow r^{2}=4$
$\Rightarrow r= \pm 2$
When $r=2, x=\dfrac{3}{1+2}=\dfrac{3}{3}=1$
When $r=-2, x=\dfrac{3}{1-2}=\dfrac{3}{-1}=-3$
Case I:
When $r=2$ and $x=1$,
$a b=x^{2} r=2$
$c d=x^{2} r^{5}=32$ $\therefore \dfrac{q+p}{q-p}=\dfrac{32+2}{32-2}=\dfrac{34}{30}=\dfrac{17}{15}$
i.e., $(q+p):(q-p)=17: 15$
Case II:
When $r=-2, x=$1,
$a b=x^{2} r=-18$
$c d=x^{2} r^{5}=- 288$
$\therefore \dfrac{q+p}{q-p}=\dfrac{-288-18}{-288+18}=\dfrac{-306}{-270}=\dfrac{17}{15}$
i.e., $(q+p):(q-p)=17: 15$
Thus, in both the cases, we obtain $(q+p):(q$ - $p)=17: 15$
10. The ratio of the A.M. and G.M. of two positive numbers $a$ and $b$, is $m: n$. Show that $a: b=(m+\sqrt{m^{2}-n^{2}}):(m-\sqrt{m^{2}-n^{2}})$.
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Answer :
Let the two numbers be $a$ and $b$.
A.M $=\dfrac{a+b}{2}$ and G.M. $=\sqrt{a b}$
According to the given condition,
$\dfrac{a+b}{2 \sqrt{a b}}=\dfrac{m}{n}$
$\Rightarrow \dfrac{(a+b)^{2}}{4(a b)}=\dfrac{m^{2}}{n^{2}}$
$\Rightarrow(a+b)^{2}=\dfrac{4 a b m^{2}}{n^{2}}$
$\Rightarrow(a+b)=\dfrac{2 \sqrt{a b} m}{n}$
Using this in the identity $(a \text{ - } b)^{2}=(a+b)^{2}$ - $4 a b$, we obtain
$$ \begin{align*} & (a-b)^{2}=\dfrac{4 a b m^{2}}{n^{2}}-4 a b=\dfrac{4 a b(m^{2}-n^{2})}{n^{2}} \\ & \Rightarrow(a-b)=\dfrac{2 \sqrt{a b} \sqrt{m^{2}-n^{2}}}{n} \tag{2} \end{align*} $$
Adding (1) and (2), we obtain
$$ \begin{aligned} & 2 a=\dfrac{2 \sqrt{a b}}{n}(m+\sqrt{m^{2}-n^{2}}) \\ & \Rightarrow a=\dfrac{\sqrt{a b}}{n}(m+\sqrt{m^{2}-n^{2}}) \end{aligned} $$
Substituting the value of $a$ in (1), we obtain
$$ \begin{aligned} & b=\dfrac{2 \sqrt{a b}}{n} m-\dfrac{\sqrt{a b}}{n}(m+\sqrt{m^{2}-n^{2}}) \\ & =\dfrac{\sqrt{a b}}{n} m-\dfrac{\sqrt{a b}}{n} \sqrt{m^{2}-n^{2}} \\ & =\dfrac{\sqrt{a b}}{n}(m-\sqrt{m^{2}-n^{2}}) \\ & \therefore a: b=\dfrac{a}{b}=\dfrac{\dfrac{\sqrt{a b}}{n}(m+\sqrt{m^{2}-n^{2}})}{\dfrac{\sqrt{a b}}{n}(m-\sqrt{m^{2}-n^{2}})}=\dfrac{(m+\sqrt{m^{2}-n^{2}})}{(m-\sqrt{m^{2}-n^{2}})} \end{aligned} $$
Thus, $a: b=(m+\sqrt{m^{2}-n^{2}}):(m-\sqrt{m^{2}-n^{2}})$
11. Find the sum of the following series up to $n$ terms:
(i) $5+55+555+\ldots$
(ii) $.6+.66+.666+\ldots$
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Answer :
(i) $5+55+555+\ldots$
Let $S_n=5+55+555+\ldots$. to $n$ terms
$ \begin{aligned} & =\dfrac{5}{9}[9+99+999+\ldots \text{ to } n \text{ terms }] \\ & =\dfrac{5}{9}[(10-1)+(10^{2}-1)+(10^{3}-1)+\ldots \text{ to } n \text{ terms }] \\ & =\dfrac{5}{9}[(10+10^{2}+10^{3}+\ldots \text{ n terms })-(1+1+\ldots n \text{ terms })] \\ & =\dfrac{5}{9}[\dfrac{10(10^{n}-1)}{10-1}-n] \\ & =\dfrac{5}{9}[\dfrac{10(10^{n}-1)}{9}-n] \\ & =\dfrac{50}{81}(10^{n}-1)-\dfrac{5 n}{9} \end{aligned} $
(ii) $.6+.66+.666+\ldots$
Let $S_n=06 .+0.66+0.666+\ldots$ to $n$ terms
$=6[0.1+0.11+0.111+\ldots$ to $n$ terms $]$
$=\dfrac{6}{9}[0.9+0.99+0.999+\ldots$ to $n$ terms $]$
$=\dfrac{6}{9}[(1-\dfrac{1}{10})+(1-\dfrac{1}{10^{2}})+(1-\dfrac{1}{10^{3}})+\ldots.$ to n terms $]$
$=\dfrac{2}{3}[(1+1+\ldots n.$ terms $)-\dfrac{1}{10}(1+\dfrac{1}{10}+\dfrac{1}{10^{2}}+\ldots n.$ terms $.)]$
$=\dfrac{2}{3}[n-\dfrac{1}{10}(\dfrac{1-(\dfrac{1}{10})^{n}}{1-\dfrac{1}{10}})]$
$=\dfrac{2}{3} n-\dfrac{2}{30} \times \dfrac{10}{9}(1-10^{-n})$
$=\dfrac{2}{3} n-\dfrac{2}{27}(1-10^{-n})$
12. Find the $20^{\text{th }}$ term of the series $2 \times 4+4 \times 6+6 \times 8+\ldots+n$ terms.
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Answer :
The given series is $2 \times 4+4 \times 6+6 \times 8+\ldots n$ terms
$\therefore n^{\text{th }}$ term $=a_n=2 n \times(2 n+2)=4 n^{2}+4 n$
$a _{20}=4(20)^{2}+4(20)=4(400)+80=1600+80=1680$
Thus, the $20^{\text{th }}$ term of the series is 1680 .
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?
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Answer :
It is given that the farmer pays Rs 6000 in cash.
Therefore, unpaid amount = Rs 12000 - Rs $6000=$ Rs 6000
According to the given condition, the interest paid annually is
$12 %$ of $6000,12 %$ of $5500,12 %$ of $5000, \ldots, 12 %$ of 500
Thus, total interest to be paid $=12 %$ of $6000+12 %$ of $5500+12 %$ of $5000+\ldots+12 %$ of 500
$=12 %$ of $(6000+5500+5000+\ldots+500)$
$=12 %$ of $(500+1000+1500+\ldots+6000)$
Now, the series $500,1000,1500 \ldots 6000$ is an A.P. with both the first term and common difference equal to 500.
Let the number of terms of the A.P. be $n$.
$\therefore 6000=500+(n$ - 1$) 500$
$\Rightarrow 1+(n- 1)=12$
$\Rightarrow n=12$
$\therefore$ Sum of the A.P
$ =\dfrac{12}{2}[2(500)+(12-1)(500)]=6[1000+5500]=6(6500)=39000 $
Thus, total interest to be paid $=12 %$ of $(500+1000+1500+\ldots+6000)$
$=12 %$ of $39000=$ Rs 4680
Thus, cost of tractor $=($ Rs $12000+$ Rs 4680) $=$ Rs 16680
14. Shamshad Ali buys a scooter for Rs 22000 . He pays Rs 4000 cash and agrees to pay the balance in annual instalment of Rs 1000 plus $10$ % interest on the unpaid amount. How much will the scooter cost him?
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Answer :
It is given that Shamshad Ali buys a scooter for Rs 22000 and pays Rs 4000 in cash.
$\therefore$ Unpaid amount $=$ Rs 22000 - Rs $4000=$ Rs 18000
According to the given condition, the interest paid annually is
$10 %$ of $18000,10 %$ of $17000,10 %$ of $16000 \ldots 10 %$ of 1000
Thus, total interest to be paid $=10 %$ of $18000+10 %$ of $17000+10 %$ of $16000+\ldots+10 %$ of 1000
$=10 %$ of $(18000+17000+16000+\ldots+1000)$
$=10 %$ of $(1000+2000+3000+\ldots+18000)$
Here, 1000, 2000, $3000 \ldots 18000$ forms an A.P. with first term and common difference both equal to 1000.
Let the number of terms be $n$.
$\therefore 18000=1000+(n$ - 1 $)(1000)$
$\Rightarrow n=18$
$ \begin{aligned} & \begin{aligned} \therefore 1000+2000+\ldots+18000 & =\dfrac{18}{2}[2(1000)+(18-1)(1000)] \\ & =9[2000+17000] \\ & =171000 \end{aligned} \\ \end{aligned} $
After, total interest of 10%, amount will be $ =\text { Rs- }17100 $
$\therefore$ Cost of the scooter is
$ =22000+17100 $
$ =\text { 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 $8^{\text{th }}$ set of letter is mailed.
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Answer :
The numbers of letters mailed forms a G.P.: $4,4^{2}, \ldots 4^{8}$
First term $=4$
Common ratio $=4$
Number of terms $=8$
It is known that the sum of $n$ terms of a G.P. is given by
$ \begin{aligned} & S_n=\dfrac{a(r^{n}-1)}{r-1} \\ & \therefore S_8=\dfrac{4(4^{8}-1)}{4-1}=\dfrac{4(65536-1)}{3}=\dfrac{4(65535)}{3}=4(21845)=87380 \end{aligned} $
It is given that the cost to mail one letter is 50 paisa.
$\therefore$ Cost of mailing 87380 letters $=\operatorname{Rs~} 87380 \times \dfrac{100}{100}=$ Rs 43690
$ =Rs 87380 \times \dfrac{50}{100}=\text{ Rs } 43690 $
Thus, the amount spent when $8^{\text{th }}$ set of letter is mailed is Rs 43690 .
16. 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.
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Answer :
It is given that the man deposited Rs 10000 in a bank at the rate of 5% simple interest annually.
$\therefore$ Interest in first year
$ =\dfrac{5}{100} \times Rs 10000=Rs 500 $
$\therefore$ Amount in $15^{\text{th }}$ year $=Rs$
$ 10000+\underbrace{500+500+\ldots+500} _{14 \text{ times }} $
$=$ Rs $10000+14 \times$ Rs 500
= Rs $10000+$ Rs 7000
= Rs 17000
Amount after 20 years $=$
$ Rs 10000+\underbrace{500+500+\ldots+500} _{20 \text{ times }} $
$=Rs 10000+20 \times Rs 500$
$=Rs 10000+Rs 10000$
= 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.
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Answer :
Cost of machine $=$ Rs 15625
Machine depreciates by 20% every year.
Therefore, its value after every year is $80 %$ of the original cost i.e., $\dfrac{4}{5}$ of the original cost.
$\therefore$ Value at the end of 5 years $=$
$ 15625 \times \underbrace{\dfrac{4}{5} \times \dfrac{4}{5} \times \ldots \times \dfrac{4}{5}} _{5 \text{ times }}=5 \times 1024=5120 $
Thus, the value of the machine at the end of 5 years is Rs 5120 .
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.
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Answer :
Let $x$ be the number of days in which 150 workers finish the work.
According to the given information,
$150 x=150+146+142+\ldots .(x+8)$ terms
The series $150+146+142+\ldots .(x+8)$ terms is an A.P. with first term 146 , common difference - 4 and number of terms as $(x+8)$
$ \begin{aligned} & \Rightarrow 150 x=\dfrac{(x+8)}{2}[2(150)+(x+8-1)(-4)] \\ & \Rightarrow 150 x=(x+8)[150+(x+7)(-2)] \\ & \Rightarrow 150 x=(x+8)(150-2 x-14) \\ & \Rightarrow 150 x=(x+8)(136-2 x) \\ & \Rightarrow 75 x=(x+8)(68-x) \\ & \Rightarrow 75 x=68 x-x^{2}+544-8 x \\ & \Rightarrow x^{2}+75 x-60 x-544=0 \\ & \Rightarrow x^{2}+15 x-544=0 \\ & \Rightarrow x^{2}+32 x-17 x-544=0 \\ & \Rightarrow x(x+32)-17(x+32)=0 \\ & \Rightarrow(x-17)(x+32)=0 \\ & \Rightarrow x=17 \text{ or } x=-32 \end{aligned} $
However, $x$ cannot be negative.
$\therefore x=17$
Therefore, originally, the number of days in which the work was completed is 17.
Thus, required number of days $=(17+8)=25$
