Mr. Nair gets ₹6455 at the end of one year at the rate of 14% per annum in a recurring deposit account. Find the monthly installment.

Here,
n = number of months for which the money is deposited = 1 x 12 = 12,
r = interest rate per annum = 14

Let the monthly installment be ₹x, then P = ₹x.

Using the formula:

$I = P \times \dfrac{n(n+1)}{2 \times 12} \times \dfrac{r}{100} \text{, we get} \\[0.7em] I = \Big( x \times \dfrac{12 \times 13}{2 \times 12} \times \dfrac{14}{100} \Big) \\[0.5em] \enspace\medspace = ₹\dfrac{91}{100}x$

Total money deposited by Mr. Nair = ₹12x

∴ The amount of maturity = total money deposited + interest

$= ₹12x + ₹\dfrac{91}{100}x \\[0.5em] = ₹\dfrac{1291}{100}x$

According to the given,

Amount of maturity = ₹6455

$\Rightarrow \dfrac{1291}{100}x = 6455 \\[0.5em] \Rightarrow \dfrac{x}{100} = \dfrac{6455}{1291} \\[0.5em] \Rightarrow x = 5 \times 100 \\[0.5em] \Rightarrow x = 500$

∴ The monthly installment = ₹500