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Q)

A man deposited $Rs. 10,000$ 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 find the total amount after $20$ ears.

$\begin{array}{1 1}Rs. 16,000 \:and\:Rs. 2,94000 \\ Rs. 17,000 \:and\:Rs. 29000 \\Rs. 16,500 \:and\:Rs. 2,95000 \\ Rs. 17,000 \:and\:Rs. 2,95000 \end{array}$

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A)
Toolbox:
• $n^{th}$ term of an A.P., $t_n=a+(n-1)d$
• Sum of $n$ terms of an A.P., $S_n=\large\frac{n}{2}$$\big[2a+(n-1)d\big] Given that the initial deposit =Rs.10,000 Interest= 5% (simple interest). Interest after one year =\large\frac{5}{100}$$\times 10000$
$\qquad\qquad\:=500$
$\therefore$ After one year the ampount $=10,000+500=10500$
Similarly after two years the amount $=10,000+2\times 500=11,000$
and so on.
This amount increases in A.P.
$\therefore$ The amount in 15$^{th}$ years is $15^{th}$ term of the series
$10,000+10,500+11,000+.......$ which is A.P. with
first term $=a=10,000$ and common difference $=500$
We know that $n^{th}$ term of an A.P., $t_n=a+(n-1)d$
$\therefore$ The amount in $15^{th}$ year $=t_{15}= 10,000+(15-1)500\big]$
$=10,000+7000=Rs.17,000$
Step 2
The total amount after $20$ years is the sum of $20$ terms of the above series.
We know that the sum of $n$ terms of an A.P., $S_n=\large\frac{n}{2}$$\big[2a+(n-1)d\big] \therefore S_{20}=\large\frac{20}{2}$$\big[2\times 10000+(20-1)500\big]$$=295000.$
$i.e.,$ Total amount after 20 years $=Rs. 2,95,000$