Question

A recurring deposit account of Rs 1,200 per month has a maturity value of Rs 12,440. 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.

Answer 1

$P=Rs.1200,noofmonths=n,rate=8\%MaturityAmount=Rs.12440$

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

$12440=1200\times n+1200\times \frac{n(n+1)}{2\times 12}\times \frac{8}{100}$

$12440=1200n+4n(n+1)$

$n^2+301n-3110=0$

$n^2+(311-10)n-3110=0$

$n^2+311n-10n-3110=0$

$n(n+311)-10(n+311)=0$

$(n+311)(n-10)=0$

$n=-311or10$

So n=10

Hence n = 10 months.