Find the compound interest on ₹3125 for 3 years if the rates of interest for the first, second and third year are respectively 4%, 5% and 6% per annum.

A = $P\Big(1 + \dfrac{r}{100}\Big)^n$

For first year, P = ₹3125 and rate = 4%.

Using formula,

$A = P\Big(1 + \dfrac{r}{100}\Big)^n \\[1em] = ₹3125 \times \Big(1 + \dfrac{4}{100}\Big)^1 \\[1em] = ₹3125 \times \Big(\dfrac{104}{100}\Big) \\[1em] = ₹3125 \times \dfrac{26}{25} \\[1em] = ₹\dfrac{81250}{25} \\[1em] = ₹3250.$

For second year, P = ₹3250 and rate = 5%.

Using formula,

$A = ₹3250 \times \Big(1 + \dfrac{5}{100}\Big)^1 \\[1em] = ₹3250 \times \Big(\dfrac{105}{100}\Big) \\[1em] = ₹3250 \times \dfrac{21}{20} \\[1em] = ₹\dfrac{68250}{20} \\[1em] = ₹3412.50$

For third year, P = ₹3412.50 and rate = 6%.

Using formula,

$A = ₹3412.50 \times \Big(1 + \dfrac{6}{100}\Big)^1 \\[1em] = ₹3412.50 \times \Big(\dfrac{106}{100}\Big) \\[1em] = ₹3412.50 \times \dfrac{53}{50} \\[1em] = ₹\dfrac{180862.5}{50} \\[1em] = ₹3617.25$

C.I. = Final Amount - Principal = ₹3617.25 - ₹3125 = ₹492.25

Hence, compound interest = ₹492.25