Mohan has a recurring deposit account in a bank for 2 years at 6% p.a. simple interest. If he gets ₹1200 as interest at the time of maturity, find

(i) the monthly installment.
(ii) the amount of maturity.

Here,
n = number of months for which the money is deposited = 2 x 12 = 24,
r = interest rate per annum = 6

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

Using the formula:

$I = P \times \dfrac{n(n+1)}{2 \times 12} \times \dfrac{r}{100} \text{, we get} \\[0.7em] I = \Big( x \times \dfrac{24 \times 25}{2 \times 12} \times \dfrac{6}{100} \Big) \\[0.5em] \enspace\medspace = ₹1.5x$

According to the given,

$1.5x = 1200 \\[0.5em] \Rightarrow x = \dfrac{1200}{1.5} \\[0.5em] \Rightarrow x = 800 \\[0.5em]$

∴ The monthly installment = ₹800

(ii) Total amount deposited by Mohan = ₹(800 x 24) = ₹19200

∴ Amount of maturity = total amount deposited + interest
= ₹19200 + ₹1200
= ₹20400