Mr. Gulati has a Recurring deposit account of ₹ 300 per month. If the rate of interest is 12% and the maturity value of this account is ₹ 8100; find the time (in years) of this recurring deposit account.

Let time of this recurring deposit be x months.

So,

P = ₹ 300, n = x months and r = 12%

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

$\therefore I = ₹ 300 \times \dfrac{x(x + 1)}{2 \times 12} \times \dfrac{12}{100} \\[1em] = \dfrac{3x(x + 1)}{2}$

Maturity value = Sum deposited + Interest

$\Rightarrow 300x + \dfrac{3x(x + 1)}{2} = 8100 \\[1em] \Rightarrow \dfrac{600x + 3x^2 + 3x}{2} = 8100 \\[1em] \Rightarrow 3x^2 + 603x = 16200 \\[1em] \Rightarrow 3x^2 + 603x - 16200 = 0 \\[1em] \Rightarrow 3(x^2 + 201x - 5400) = 0 \\[1em] \Rightarrow x^2 + 201x - 5400 = 0 \\[1em] \Rightarrow x^2 + 225x - 24x - 5400 = 0 \\[1em] \Rightarrow x(x + 225) - 24(x + 225) = 0 \\[1em] \Rightarrow (x - 24)(x + 225) = 0 \\[1em] \Rightarrow x = 24 \text{ or } x = -225.$

Since, time cannot be negative.

∴ x = 24 months or 2 years.

Hence, the time of this recurring deposit account is 2 years.