A man deposits ₹ 1800 per month in a recurring deposit account for 2 years. If he gets ₹ 3600 as interest at the time of maturity, find the rate of interest.

Given,

P = ₹ 1800

Time (n) = 2 years = 24 months

Interest = ₹ 3600

Let rate of interest be r%.

By formula,

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

Substituting values we get :

$\Rightarrow 3600 = 1800 \times \dfrac{24 \times 25}{2 \times 12} \times \dfrac{r}{100} \\[1em] \Rightarrow 3600 = 1800 \times \dfrac{r}{4} \\[1em] \Rightarrow r = \dfrac{3600 \times 4}{1800} \\[1em] \Rightarrow r = 8 \%.$

Hence, rate of interest = 8%.