Find the sum invested at 10% compounded annually, on which the interest for the third year, exceeds the interest of the first year by ₹ 252.

Let 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 for the third year exceeds the interest of the first year by ₹ 252.

$\therefore \dfrac{121x}{1000} - \dfrac{x}{10} = 252 \\[1em] \Rightarrow \dfrac{121x - 100x}{1000} = 252 \\[1em] \Rightarrow \dfrac{21x}{1000} = 252 \\[1em] \Rightarrow x = \dfrac{252 \times 1000}{21} = 12000.$

Hence, sum = ₹ 12000.