Suresh has a recurring deposit account in a bank. He deposits ₹2000 per month and the bank pays interest at the rate of 8% per annum. If he gets ₹1040 as interest at the time of maturity, find in years the total time for which the account was held.
Monthly deposit (P) = ₹2000
Rate of interest (r) = 8% per annum
Interest (I) = ₹1040
Let the account be held for n months.
For a recurring deposit:
I = P × \frac{n(n+1)}{2}\times\frac{r}{12\times100}
Substituting the values:
1040 = 2000\times\frac{n(n+1)}{2}\times\frac{8}{1200}
⇒ 1040 = \frac{2000n(n+1)}{300}
⇒ n(n + 1) = \frac{1040\times300}{2000}
⇒ n(n + 1) = 156
⇒ n² + n − 156 = 0
⇒ n² + 13n − 12n − 156 = 0
⇒ n(n + 13) − 12(n + 13) = 0
⇒ (n − 12)(n + 13) = 0
⇒ n = 12 or n = −13
The number of months cannot be negative, so n = 12 months.
12 months = 1 year.
Therefore, the account was held for 1 year.
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