A person deposited ₹ $ 400$ at the beginning of every month in a recurring deposit account and received ₹ $ 16,398$ at the end of $3$ years. Find the rate of interest given by the bank.
Money deposited each month $= P =$ ₹ $ 400$
Time for which money deposited $= n = 3 $ years $= 36 $ months
Let, rate of interest $= r \%$
Amount received on maturity $= $ ₹ $ 16,398$
$\begin{aligned}
\text{Interest} & = P \times \dfrac{n \left(n + 1\right)}{2 \times 12} \times \dfrac{r}{100} \\\\
& = 400 \times \dfrac{36 \times 37}{2 \times 12} \times \dfrac{r}{100} \\\\
& = 222 \; r
\end{aligned}$
i.e. $\;$ Interest $= $ ₹ $ 222 \; r$
Total money deposited $= 400 \times 36 = $ ₹ $ 14,400$
$\text{Amount received on maturity} = \text{Money deposited} + \text{Interest}$
i.e. $\;$ $16398 = 14400 + 222 \; r$
i.e. $\;$ $222 \; r = 1998$
i.e. $\;$ $r = 9$
$\therefore \;$ Rate of interest given by the bank $= 9 \%$