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

Let time = n months.

Given,

P = ₹ 2000

r = 10%

By formula,

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

Substituting values we get :

$\Rightarrow 83100 = 2000 \times n + 2000 \times \dfrac{n(n + 1)}{2 \times 12} \times \dfrac{10}{100} \\[1em] \Rightarrow 83100 = 2000n + \dfrac{100n(n + 1)}{12} \\[1em] \Rightarrow 83100 = \dfrac{24000n + 100n^2 + 100n}{12} \\[1em] \Rightarrow 83100 \times 12 = 100(240n + n^2 + n) \\[1em] \Rightarrow 241n + n^2 = \dfrac{83100 \times 12}{100} \\[1em] \Rightarrow 241n + n^2 = 9972 \\[1em] \Rightarrow n^2 + 241n - 9972 = 0 \\[1em] \Rightarrow n^2 + 277n - 36n - 9972 = 0\\[1em] \Rightarrow n(n + 277) - 36(n + 277) = 0 \\[1em] \Rightarrow (n - 36)(n + 277) = 0 \\[1em] \Rightarrow n - 36 = 0 \text{ or } n + 277 = 0 \\[1em] \Rightarrow n = 36 \text{ or } n = -277.$

Since, time cannot be negative.

∴ n = 36 months = 3 years.

Hence, time = 3 years.