David opened a Recurring Deposit Account in a bank and deposited ₹300 per month for two years. If he received ₹7725 at the time of maturity, find the rate of interest.

Here,
P = money deposited per month = ₹300,
n = number of months for which the money is deposited = 2 x 12 = 24

Let the rate of interest be r% per annum, then by using the formula:

$I = P \times \dfrac{n(n+1)}{2 \times 12} \times \dfrac{r}{100} \text{, we get} \\[0.7em] I = \Big( 300 \times \dfrac{24 \times 25}{2 \times 12} \times \dfrac{r}{100} \Big) \\[0.5em] \enspace\medspace = 75r$

Total money deposited by David = ₹300 x 24 = ₹7200

∴ The amount of maturity = total money deposited + interest
= 7200 + 75r

According to the given,

$7200 + 75r = 7725 \\[0.5em] \Rightarrow 75r = 7725 - 7200 \\[0.5em] \Rightarrow 75r = 525 \\[0.5em] \Rightarrow r = \dfrac{525}{75} \\[0.5em] \Rightarrow r = 7$

∴ Rate of (simple) interest = 7% p.a.