Question

14. Arnav deposits 500 every month at 12% p.a.
in a bank in a recurring deposit scheme. The
bank pays 5275 on maturity. Find the time
for which the account is held.
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Answer 1

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Answer 2

Arnav's account was held for 10 months.

Step-by-step explanation:

Let the account is held for 'n' months.

The amount of monthly deposit (P) = 500

Rate of interest (r) = 12%

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

            $=500\times \frac{n(n+1)}{2\times 12}\times \frac{12}{100}$

            $=\frac{5n(n+1)}{2}$

Maturity value = deposited amount + interest

$5275=500n+\frac{5n(n+1)}{2}$

$5275=\frac{5}{2}n^2+1005n^2$

$\frac{-5}{2}n^2+\frac{-1005}{2}n+5275=0$

Use quadratic formula for a = -2.5 b = -502.5 c = 5275

$n=\frac{-(-502.5)\pm\sqrt{(-502.5)^2-4(-2.5)(5275)} }{2(-2.5)}$

$n=\frac{(502.5)\pm\sqrt{305256.25} }{-5}$

n = -211, 10

months can not be in negative form, so n = 10 months

Arnav's account was held for 10 months.

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