Find the compound interest to the nearest rupee on ₹ 10800 for $2\dfrac{1}{2}$ years at 10% per annum.

Given,

P = ₹ 10800

T = 2.5 years

r = 10%

For first 2 years :

$A = P\Big(1 + \dfrac{r}{100}\Big)^n \\[1em] = 10800 \times \Big(1 + \dfrac{10}{100}\Big)^2 \\[1em] = 10800 \times \Big(\dfrac{110}{100}\Big)^2\\[1em] = 10800 \times \Big(\dfrac{11}{10}\Big)^2 \\[1em] = 10800 \times \dfrac{121}{100} \\[1em] = ₹ 13068.$

For next $\dfrac{1}{2}$ year :

P = ₹ 13068

$A = P\Big(1 + \dfrac{r}{2 \times 100}\Big)^{n \times 2} \\[1em] = 13068 \times \Big(1 + \dfrac{10}{2 \times 100}\Big)^{\dfrac{1}{2} \times 2} \\[1em] = 13068 \times \Big(1 + \dfrac{1}{20}\Big)^1 \\[1em] = 13068 \times \dfrac{21}{20} \\[1em] = ₹ 13721.$

By formula,

C.I. = A - P = ₹ 13721 - ₹ 10800 = ₹ 2921.

Hence, compound interest = ₹ 2921.