The maturity value of a R.D. Account is ₹3,320. If the monthly instalment is ₹400 and the rate of interest is 10%; find the time (period) of this R.D. Account.

Given, P = ₹400, r = 10% and MV = ₹3,320

Let n be the number of months.

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

$\therefore I = ₹ 400 \times \dfrac{n \times (n + 1)}{2 \times 12} \times \dfrac{10}{100} \\[1em] = ₹ 400 \times \dfrac{n(n + 1)}{24} \times \dfrac{1}{10} \\[1em] = ₹ \dfrac{40n(n + 1)}{24} \\[1em] = ₹ \dfrac{5n(n + 1)}{3}$

Maturity value = Sum deposited + Interest

$\Rightarrow ₹ 3,320 = ₹ 400 \times n + ₹ \dfrac{5n(n + 1)}{3}\\[1em] \Rightarrow 3,320 = 400n + \dfrac{5n^2 + 5n}{3}\\[1em] \Rightarrow 3,320 = \dfrac{1200n + 5n^2 + 5n}{3}\\[1em] \Rightarrow 3,320 \times 3 = 1200n + 5n^2 + 5n\\[1em] \Rightarrow 9,960 = 1200n + 5n^2 + 5n\\[1em] \Rightarrow 1200n + 5n^2 + 5n - 9,960 = 0\\[1em] \Rightarrow 5n^2 + 1205n - 9,960 = 0\\[1em] \Rightarrow n^2 + 241n - 1992 = 0\\[1em] \Rightarrow n^2 + 249n - 8n - 1992 = 0\\[1em] \Rightarrow n(n + 249) - 8(n + 249) = 0\\[1em] \Rightarrow (n + 249)(n - 8) = 0\\[1em] \Rightarrow (n + 249) = 0 \text{ or } (n - 8) = 0\\[1em] \Rightarrow n = -249 \text{ or } n = 8\\[1em]$

Number of months cannot be negative.

Hence, number of months = 8.