Suresh has a recurring deposit account in a bank. He deposits ₹2000 per month and the bank pays interest at the rate of 8% per annum. If he gets ₹1040 as interest at the time of maturity, find in years total time for which the account was held.
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Let time be n months.
By formula,
I = $P \times \frac{n(n + 1)}{2\times 12} \times \frac{r}{100}$
Substituting values we get :
$$\begin{aligned}\Rightarrow 1040 = 2000 \times \frac{n(n + 1)}{24} \times \frac{8}{100} \\[1em] \Rightarrow 1040 = 2000 \times \frac{n(n + 1)}{300} \\[1em] \Rightarrow n(n + 1) = \frac{1040 \times 300}{2000} \\[1em] \Rightarrow n(n + 1) = 156 \\[1em] \Rightarrow n^2 + n - 156 = 0 \\[1em] \Rightarrow n^2 + 13n - 12n - 156 = 0 \\[1em] \Rightarrow n(n + 13) - 12(n + 13) = 0 \\[1em] \Rightarrow (n - 12)(n + 13) = 0 \\[1em] \Rightarrow n - 12 = 0 \text{ or } n + 13 = 0 \\[1em] \Rightarrow n = 12 \text{ or } n = -13.\end{aligned}$$
Since, no. of months cannot be negative.
∴ n = 12.
Hence, total time for which the account was held = 12 months.