Calculate the compound interest for the second year on ₹ 12,000 invested for 3 years at 10% per year. Also, find the sum due at the end of the third year.

For the first year:

P = ₹ 12,000, R = 10 %, T = 1 year

$\text{Interest} = \dfrac{P \times R \times T}{100}\\[1em] = \dfrac{12,000 \times 10 \times 1}{100}\\[1em] = \dfrac{120,000}{100}\\[1em] = ₹ 1,200$

Amount at the end of first year = P + I

= ₹ 12,000 + 1,200

= ₹ 13,200

For the second year:

P = ₹ 13,200, R = 10 %, T = 1 year

$\text{Interest} = \dfrac{13,200 \times 10 \times 1}{100}\\[1em] = \dfrac{132,000}{100}\\[1em] = ₹ 1,320$

Amount at the end of second year = P + I

= ₹ 13,200 + 1,320

= ₹ 14,520

For the third year:

P = ₹ 14,520, R = 10 %, T = 1 year

$\text{Interest} = \dfrac{14,520 \times 10 \times 1}{100}\\[1em] = \dfrac{145,200}{100}\\[1em] = ₹ 1,452$

Amount at the end of third year = P + I

= ₹ 14,520 + 1,452

= ₹ 15,972

Hence, the compound interest for the second year = ₹ 1,320 and the sum due at the end of the third year = ₹ 15,972.