Rajiv Bhardwaj has a recurring deposit account in a bank of ₹600 per month. If the bank pays simple interest of 7% p.a. and he gets ₹15450 as maturity amount, find the total time for which the account was held.

Here,
P = money deposited per month = ₹600,
r = simple interest rate percent per annum = 7

Let the account be held for n months

Using the formula:

$I = P \times \dfrac{n(n+1)}{2 \times 12} \times \dfrac{r}{100} \text{, we get} \\[0.7em] I = \Big( 600 \times \dfrac{n(n+1)}{2 \times 12} \times \dfrac{7}{100} \Big) \\[0.5em] \enspace\medspace = \dfrac{7n(n+1)}{4}$

Total money deposited by Rajiv Bhardwaj = ₹(600 x n) = ₹600n

∴ Amount of maturity = total amount deposited + interest

$= ₹600n + \dfrac{7n(n+1)}{4} \\[0.5em] = ₹\dfrac{2400n + 7n(n+1)}{4} \\[0.5em] = ₹\dfrac{7n^2 + 2407n}{4}$

According to the given,

$\dfrac{7n^2 + 2407n}{4} = 15450 \\[0.5em] \Rightarrow 7n^2 + 2407n - 61800 = 0 \\[0.5em] \Rightarrow 7n(n - 24) + 2575(n - 24) = 0 \\[0.5em] \Rightarrow (n - 24)(7n + 2575) = 0 \\[0.5em] \Rightarrow n = 24, -\dfrac{2575}{7} \\[0.5em] \text{ (but n cannot be negative)} \\[0.5em] \Rightarrow n = 24$

∴ The account was held for 24 months i.e. 2 years.