Divya deposited $R s .1000$ at compound interest at the rate of $10 %$ per annum. The amount at the end of first year, second year, third year,.. and so on.

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Solution

$A = p {[1 + \frac{r}{100}]}^{t} p = 1000 R s . A = A m o u n t .$

$t = 1 y e a r \Rightarrow A = 1000 [1 + \frac{10}{100}]^{1} = 1100$

$t = 2 y e a r \Rightarrow A = 1000 [1 + \frac{10}{100}]^{2} = 1210.$

$t = n y e a r \Rightarrow A = 1000 [1.1]^{n}$

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