The monthly installment of a recurring deposit account is ₹ 2400. If the account is held for 1 year 6 months and its maturity value is ₹ 47,304, find the rate of interest.

Given,

Time (n) = 1 year 6 months = 12 + 6 = 18 months.

Monthly installment (P) = ₹ 2400.

Maturity value = ₹ 47304.

Let rate of interest be r%.

By formula,

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

Substituting values we get :

$\Rightarrow 47304 = 2400 \times 18 + 2400 \times \dfrac{18 \times 19}{2 \times 12} \times \dfrac{r}{100} \\[1em] \Rightarrow 47304 = 43200 + 2400 \times \dfrac{3 \times 19}{2 \times 2} \times \dfrac{r}{100} \\[1em] \Rightarrow 47304 - 43200 = 2400 \times \dfrac{57}{4} \times \dfrac{r}{100} \\[1em] \Rightarrow 4104 = 6 \times 57r \\[1em] \Rightarrow r = \dfrac{4104}{6 \times 57} \\[1em] \Rightarrow r = \dfrac{4104}{342} = 12 \%.$

Hence, rate of interest = 12%.