A person has a recurring deposit account of ₹ $400$ per month at $10 \%$ per annum. If the person gets ₹ $260$ as interest at the time of maturity, find the total time for which the account was held.
Let the account be held for $n$ months.
Money deposited per month $= P =$ ₹ $400$
$\therefore \;$ Total money deposited $= $ ₹ $\left(400 \times n\right)$
Rate of interest $= r = 10\%$
Interest $= I = P \times \dfrac{n \left(n + 1\right)}{2 \times 12} \times \dfrac{r}{100}$
i.e. $\;$ $I = $ ₹ $\left[400 \times \dfrac{n \left(n + 1\right)}{2 \times 12} \times \dfrac{10}{100}\right]$
i.e. $\;$ $I = $ ₹ $\left[\dfrac{5 n \left(n + 1\right)}{3}\right]$
But, interest earned $= $ ₹ $260$
$\therefore \;$ $\dfrac{5n \left(n + 1\right)}{3} = 260$
i.e. $\;$ $5n^2 + 5n = 780$
i.e. $\;$ $n^2 + n = 156$
i.e. $\;$ $n^2 + n - 156 = 0$
i.e. $\;$ $n^2 + 13 n - 12n - 156 = 0$
i.e. $\;$ $n \left(n + 13\right) - 12 \left(n + 13\right) = 0$
i.e. $\;$ $\left(n + 13\right) \left(n - 12\right) = 0$
i.e. $\;$ $n + 13 = 0$ $\;$ or $\;$ $n - 12 = 0$
i.e. $\;$ $n = -13$ $\;$ or $\;$ $n = 12$
Since time cannot be negative, the time for which the account was held $= n = 12$ months $= 1$ year