Mr. Sonu has a recurring deposit account and deposits ₹750 per month for 2 years. If he gets ₹19,125 at the time of maturity, find the rate of interest.

Given,

P = ₹750

n = 2 years = 24 months

Maturity Value = ₹19,125

Let the rate of interest be 'r' per annum

Sum deposited = P × n = 750 × 24 = ₹18,000

Maturity Value = Sum deposited + Interest

Interest = Maturity Value - Sum deposited

∴ I = ₹19,125 - ₹18,000 = ₹1,125

I = $P \times \dfrac{n(n+1)}{2 \times 12} \times \dfrac{r}{100}$

$1125 = 750 \times \dfrac{24 \times 25}{24} \times \dfrac{r}{100}\\[1em] 1125 = 750 \times 25 \times \dfrac{r}{100}\\[1em] 1125 = 18750 \times \dfrac{r}{100}\\[1em] 1125 = 187.5r\\[1em] r = \dfrac{1125}{187.5}\\[1em] r=6\%$

Hence, the rate of interest is 6% per annum.