Question

I want the answer of this question in details
the question is "Samita has a recurring deposit account in a bank of Rs.2000 per monthat the rate of 10% per annum. If she gets Rs,83100 at the time of maturity, find the total time for which the account was held.

Answer 1

M = ( R * [(1+r)n - 1 ] ) / (1-(1+r)-1/3) 
[ M - maturity amount, R = monthly installment, r=interest rate/400 , n= number of quarters]
83100 = 2000*(1+10)^n -1/ (1-(1+10)^-1/3)
find the value of n from here. 

Answer 2

The total time for which the account was held is 3 years or 36 months.

Given:

Deposited amount per amount = Rs.2000

Rate of interest = 10%

Maturity value = Rs.83,100

To find:

The total time for which the account was held.

Solution:

Let the amount be held for n months.

Now, the amount deposited per month = Rs.2000

Rate, r = 10%

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

$=2000 \times \frac{n(n+1)}{2 \times 12} \times \frac{10}{100}$

$=\frac{25 n(n+1)}{3}$

Now, maturity value = amount deposited + interest

$83100=2000 n+\frac{25 n(n+1)}{3}$

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

$249300=25 n^{2}+6025 n$

$25 n^{2}+6025 n-249300=0$

$\div \ by \ 25 \ the \ above \ equation,$

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

$n^{2}-36 n+277 n-9972=0$

$n(n-36)+277(n-36)=0$

$(n-36)(n+277)=0$

$(n-36)=0 \text { or }(n+277)=0$

$n=36 \text { or } n=-277(\text { rejected })$

Then, the time for the amount held is 36 months or 3 years.