$\begin{array}{1 1}Rs.3125 \\ Rs.5,120 \\ Rs.6,300 \\ Rs.5125 \end{array} $

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- $n^{th}$ term of a G.P. $=a.r^{n-1}$

Given that the cost of the machine $=Rs.15,625$.

The depreciation is $20$%. per year.

$\Rightarrow\:$ After one year the cost will be $15,625-\large\frac{20}{100}$$\times 15625$

$\qquad\qquad=15,625\big(1-\large\frac{1}{5}\big)$$=15,625\times(\large\frac{4}{5})$$=12,500$

After two years the cost will be $12,500-\large\frac{20}{100}$$\times 12,500$$=12,500.\big.(1-\large\frac{1}{5}\big)$

$\qquad\qquad\:=15,625\times\big(1-\large\frac{1}{5}\big)^2$

The cost after $5$ years will be $5^{th}$ term of the series

$15,625\times\big(\large\frac{4}{5}\big)$ $,15,625\times\big(\large\frac{4}{5}\big)^2,.........$ which is a G.P. with

first term $a=15,625\times \large\frac{4}{5}$ and common ratio $r=(1-\large\frac{1}{5})=\large\frac{4}{5}$

We know that $n^{th}$ term of a G..P. $=a.r^{n-1}$

$\therefore t_5=15,625\times\large\frac{4}{5}\times (\large\frac{4}{5})^4$$=5120$

$i.e.,$ The cost of the machine after $5$ years will be $Rs.5,120$

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