Preeti has a recurring deposit account of ₹1,000 per month at 10% per annum. If she gets ₹5,550 as interest at the time of maturity, find the total time for which the account was held.

Given,

Monthly deposit (P) = ₹1,000

Rate of interest (r) = 10% per annum

Interest (I) = ₹5,550

Let the total time be 'n' months.

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

$\therefore 5550 = 1000 \times \dfrac{n(n+1)}{24} \times \dfrac{10}{100}\\[1em] 5550 = 1000 \times \dfrac{10n(n+1)}{2400}\\[1em] 5550 = \dfrac{10000n(n+1)}{2400} \\[1em] 5550 = \dfrac{100n(n+1)}{24} \\[1em] 5550 = \dfrac{25n(n+1)}{6} \\[1em] 33300 = 25n(n+1)\\[1em] \dfrac{33300}{25}= n(n+1)\\[1em] n^2 + n =1332 \\[1em] n^2 + n -1332=0$

Solving quadratic equation:

$\Rightarrow n^2 + 37n - 36n- 1332 = 0\\[1em] \Rightarrow n(n + 37) - 36(n + 37) = 0\\[1em] \Rightarrow (n - 36)(n + 37) = 0\\[1em] \Rightarrow n = 36 \text{ or } n = -37$

Since the number of months cannot be negative, n = 36 months.

∴ Number of years the account was held = $\dfrac{36}{12}$ = 3 years.

Hence, the account was held for 3 years.