Suhani has a recurring deposit account in a bank of ₹2000 per month at the rate of 10% p.a. If she gets ₹83100 at the time of maturity, find the total time for which the account was held.

Here,
P = money deposited per month = ₹2000,
r = simple interest rate percent per annum = 10

Let the account be held for n months

Using the formula:

$I = P \times \dfrac{n(n+1)}{2 \times 12} \times \dfrac{r}{100} \text{, we get} \\[0.7em] I = \Big( 2000 \times \dfrac{n(n+1)}{2 \times 12} \times \dfrac{10}{100} \Big) \\[0.5em] \enspace\medspace = \dfrac{25n(n+1)}{3}$

Total money deposited by Suhani = ₹(2000 x n) = ₹2000n

∴ Amount of maturity = total amount deposited + interest

$= ₹2000n + \dfrac{25n(n+1)}{3} \\[0.5em] = ₹\dfrac{6000n + 25n(n+1)}{3} \\[0.5em] = ₹\dfrac{25n^2 + 6025n}{3}$

According to the given,

$\dfrac{25n^2 + 6025n}{3} = 83100 \\[0.5em] \Rightarrow 25n^2 + 6025n - 249300 = 0 \\[0.5em] \Rightarrow n^2 + 241n - 9972 = 0 \\[0.5em] \Rightarrow n^2 + 277n -36n - 9972 = 0 \\[0.5em] \Rightarrow n(n + 277) -36(n + 277) = 0 \\[0.5em] \Rightarrow (n + 277)(n - 36) = 0 \\[0.5em] \Rightarrow n = -277, 36 \\[0.5em] \text{ (but n cannot be negative)} \\[0.5em] \Rightarrow n = 36$

∴ The account was held for 36 months i.e. 3 years.