Question

A recurring deposit account of $Rs.1,200$ per month has maturity value of $Rs.12,440$.If the rate of interest is $8\%$ and the interest is calculated at end of every month ;find the time (in months) of this Recurring Deposit Account.

Answer 1

Installment per month = 1200

No. of months = u

Rate of Interest = 8% per annum

Simple Interest $=P\times \left(\frac{u(u+1)}{2\times 12}\right)\times \frac{r}{100}$

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

$=u(u+1)\times 4$

$=uu(u+1)$

$SI=u{u}^2+4n$

Maturity Value =

$(1200\times u)+4u^2+4u$

$12440=1204u+4u^2$

$12440=4(301u+u^2)$

$u^2+301-3110=0$

$u^2+311u-104-3110=0$

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

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

$u=-311$

$u=10$

Time can not be negative

so $u=10$

the No of months = 10