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(C) Let the principal be $P$ and the rate of interest be $R \%$. The amount after $n$ years is given by $A = P(1 + \frac{R}{100})^n$. For $2$ years: $

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$2500$

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(C) Let the principal be $P$ and the rate of interest be $R \%$. The amount after $n$ years is given by $A = P(1 + \frac{R}{100})^n$. For $2$ years: $2809 = P(1 + \frac{R}{100})^2$  --- $(i)$ For $3$ years: $2977.54 = P(1 + \frac{R}{100})^3$  --- $(ii)$ Dividing $(ii)$ by $(i)$: $\frac{2977.54}{2809} = 1 + \frac{R}{100}$ $1.06 = 1 + \frac{R}{100}$ $\frac{R}{100} = 0.06 \implies R = 6 \%$. Now,substitute $R = 6$ in equation $(i)$: $2809 = P(1 + \frac{6}{100})^2$ $2809 = P(1.06)^2$ $2809 = P(1.1236)$ $P = \frac{2809}{1.1236} = 2500$. Thus,the rate of interest is $6 \%$ and the original sum is ₹ $2500$.

$A$ certain sum at compound interest amounts in two years to ₹ $2809$ and in three years to ₹ $2977.54$. Find the rate of interest and the original sum (in ₹).

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