A recurring deposit account is opened with Dena Bank, Meerut Cantt. For this ₹ 2,000 per month (at 10% p.a.) is deposited in the bank. If the maturity value is ₹ 25,300, find the total time for which account was held.

Given,

P = ₹ 2,000

r = 10%

Maturity value = ₹ 25,300

Let 'n' be number of months for which the money is deposited.

By formula,

Interest = $P \times \dfrac{n(n + 1)}{2 \times 12} \times \dfrac{r}{100}$

Substituting values, we get :

$\Rightarrow I = 2000 \times \dfrac{n(n + 1)}{2 \times 12} \times \dfrac{10}{100} \\[1em] = 2000 \times \dfrac{n(n + 1)}{240} \\[1em] = \dfrac{50n(n + 1)}{6}.$

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

∴ Maturity value = Total amount deposited + Interest

$= 2000n + \dfrac{50n(n + 1)}{6} \\[1em] = \dfrac{12000n + 50n(n + 1)}{6} \\[1em] = \dfrac{12000n + 50n^2 + 50n}{6} \\[1em] = \dfrac{12050n + 50n^2}{6}.$

Since, maturity value = ₹ 25,300

$\Rightarrow \dfrac{12050n + 50n^2}{6} = 25300 \\[1em] \Rightarrow 12050n + 50n^2 = 25300 \times 6 \\[1em] \Rightarrow 12050n + 50n^2 = 151800 \\[1em] \Rightarrow 50n^2 + 12050n - 151800 = 0 \\[1em] \Rightarrow 50(n^2 + 241n - 3036) = 0 \\[1em] \Rightarrow n^2 + 241n - 3036 = 0 \\[1em] \Rightarrow n^2 + 253n - 12n - 3036 = 0 \\[1em] \Rightarrow n(n + 253) - 12(n + 253) = 0 \\[1em] \Rightarrow (n - 12)(n + 253) = 0 \\[1em] \Rightarrow (n - 12) = 0 \text{ or } (n + 253) = 0 \\[1em] \Rightarrow n = 12 \text{ or } n = -253$

Time cannot be negative, so we take:

n = 12 months = 1 year

Hence, the RD account was held for 1 year.