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Home  >>  CBSE XI  >>  Math  >>  Sequences and Series
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A manufacturer reckons that the value of a machine which costs him $Rs.15,625$, will depreciate each year by $20\%$. find the estimated value at the end of $5$ years.

$\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 $Rs.5,120$
answered Apr 11, 2014 by rvidyagovindarajan_1
edited Apr 11, 2014 by rvidyagovindarajan_1
 

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