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.

#### Solution

Installment per month(P) = Rs 1,200

Number of months(n) = n

Let rate of interest(r) = 8% p.a.

`:. "S.I" = "P" xx ("n"("n" +1))/(2 xx 12) xx "r"/100`

`= 1200 xx ("n"("n" + 1))/(2 xx 12) xx 8/100`

`= 1200 xx ("n"("n" + 1))/24 xx8/100 = "Rs" 4"n"("n" + 1)`

Maturity value = ₹ (1,200 × n) + ₹ 4n(n+1) = Rs (1200n+4n^{2}+4n)

Given maturity value= ₹ 12,440

Then 1200n + 4n^{2 }+ 4n = 12,440

`=> 4"n"^2 + 1204"n" - 12440 = 0`

`=> "n"^2 + 301n - 3110 = 0`

`=> ("n" + 311)("n" - 10) = 0`

=> n = -311 or n = 10 months

Then number of months = 10