A recurring deposit account of ₹ 1200 per month has a maturity value of ₹ 12440. If the rate of interest is 8% and the interest is calculated at the end of every month; find the time (in months) of this Recurring Deposit Account.

Let time of this recurring deposit be x months.

So,

P = ₹ 1200, n = x months and r = 8%

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

$\therefore I = ₹ 1200 \times \dfrac{x(x + 1)}{2 \times 12} \times \dfrac{8}{100} \\[1em] = 4x(x + 1)$

Maturity value = Sum deposited + Interest

⇒ 1200x + 4x(x + 1) = ₹ 12440

⇒ 1200x + 4x² + 4x = 12440

⇒ 4x² + 1204x = 12440

⇒ 4x² + 1204x - 12440 = 0

⇒ 4(x² + 301x - 3110) = 0

⇒ x² + 301x - 3110 = 0

⇒ x² + 311x - 10x - 3110 = 0

⇒ x(x + 311) - 10(x + 311) = 0

⇒ (x - 10)(x + 311) = 0

⇒ x = 10 or x = -311.

Since months cannot be negative.

∴ x = 10.

Hence, the time of this recurring deposit account is 10 months.