Question

A man opened a recurring deposit account with a bank and deposited Rs 250 per month. If the rate of interest is 12% per annum and the bank pays Rs 10665 on maturity, find the time for which he held the account.

Answer 1

Given : P = Rs 250, r = 12% p.a. and maturity
value = Rs 10665
Let the account be held for n months
Then, money deposited for n months
$=Rs(250\times n)$
Equivalent principal for 1 month,
$x=P\left[\frac{n(n+1)}{2}\right]=Rs[250\times \frac{n(n+1)}{2}]=Rs\left[125n\right(n+1\left)\right]$
$\therefore Interest=x\times \frac{r}{100}\times \frac{1}{12}=Rs\left[125n\right(n+1)\times \frac{12}{100}\times \frac{1}{12}]=Rs\frac{5n(n+1)}{4}$
$\therefore$ Maturity value = Sum deposited + Interest
$\Rightarrow 10665=250n+\frac{5n(n+1)}{4}\Rightarrow 42660=1000n+5n(n+1)\Rightarrow n(n+1)+200n=8532\Rightarrow n^2+201n-8532=0\Rightarrow n^2+237n-36n-8532=0\Rightarrow n(n+237)-36(n+237)=0\Rightarrow (n+237)(n-36)=0\Rightarrow (n+237)(n-36)=0\Rightarrow n+237=0orn-36=0\Rightarrow n=-237orn=36$
As the time (n) cannot be negative, so $n=36$ months or $3$ years.
Hence, the required time is 3 years