Question

Samita has a recurring deposit account in a bank of Rs2000 per month at the rate of 10% p.a.If she gets Rs83100 at the time of maturity, find the total time for which the account was held.

Answer 1

$P=Rs.2000$

$R=10$%

$M.V=Rs.83100$

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

$\therefore I=25\times \frac{n(n+1)}{24}\times \frac{10}{100}$

$=\frac{25(n^2+n)}{3}$

$MV=P\times n+I$

$\Rightarrow 83100=(2000\times n)+\frac{25}{3}{n}^2+\frac{25}{3}n$

$\Rightarrow 83100=\frac{6000n+25n^2+25n}{3}$

$\Rightarrow 83100\times 3=6000n+25n^2+25n$

$\Rightarrow 25n^2+6025n-249300=0$

Dividing by $\left(25\right)$,

$n^2+241n-9972=0$

$\Rightarrow (n-36)(n+272)=0$

$\Rightarrow n=36,orn=-272$

But ${}^{\prime }{n}^{\prime }$ can not be negative.

$\therefore n=36$