Om has 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.

Let the rate of interest be r%.

Given,

P = ₹ 750/month

n = 2 years or 24 months

M.V. = ₹ 19,125

By formula,

M.V. = P x n + P x $\dfrac{\text{n(n + 1)}}{2 \times 12} \times \dfrac{\text{r}}{100}$

Substituting values we get :

$\Rightarrow 19125 = 750 \times 24 + 750 \times \dfrac{24(24 + 1)}{2 \times 12} \times \dfrac{\text{r}}{100}\\[1em] \Rightarrow 19125 = 18000 + 750 \times \dfrac{24 \times 25}{24} \times \dfrac{\text{r}}{100}\\[1em] \Rightarrow 19125 = 18000 + 750 \times 25 \times \dfrac{\text{r}}{100}\\[1em] \Rightarrow 19125 = 18000 + 750 \times \dfrac{\text{r}}{4}\\[1em] \Rightarrow 750 \times \dfrac{\text{r}}{4} = 19125 - 18000 \\[1em] \Rightarrow 750 \times \dfrac{\text{r}}{4} = 1125\\[1em] \Rightarrow \text{r} = \dfrac{1125 \times 4}{750}\\[1em] \Rightarrow \text{r} = \dfrac{4500}{750} \\[1em] \Rightarrow \text{r} = 6\%$

Hence, rate of interest = 6%.