On a certain sum of money, invested at the rate of 10 percent per annum compounded annually, the interest for the first year plus the interest for the third year is ₹ 2,652. Find the sum.

Let the sum of money be ₹ x.

For first year :

P = ₹ x

R = 10%

T = 1 year

I = $\dfrac{P \times R \times T}{100} = \dfrac{x \times 10 \times 1}{100} = \dfrac{x}{10}$.

Amount = P + I = x + $\dfrac{x}{10} = \dfrac{11x}{10}$.

For second year :

P = ₹ $\dfrac{11x}{10}$

R = 10%

T = 1 year

I = $\dfrac{P \times R \times T}{100} = \dfrac{\dfrac{11x}{10} \times 10 \times 1}{100} = \dfrac{11x}{100}$.

Amount = P + I = $\dfrac{11x}{10} + \dfrac{11x}{100} = \dfrac{110x + 11x}{100} = \dfrac{121x}{100}$.

For third year :

P = ₹ $\dfrac{121x}{100}$

R = 10%

T = 1 year

I = $\dfrac{P \times R \times T}{100} = \dfrac{\dfrac{121x}{100} \times 10 \times 1}{100} = \dfrac{121x}{1000}$.

Given, interest of first year plus the interest of third year is ₹ 2652.

$\Rightarrow \dfrac{x}{10} + \dfrac{121x}{1000} = 2652 \\[1em] \Rightarrow \dfrac{100x + 121x}{1000} = 2652 \\[1em] \Rightarrow \dfrac{221x}{1000} = 2652 \\[1em] \Rightarrow x = \dfrac{2652 \times 1000}{221} = 12000.$

Hence, sum of money is ₹ 12000.