A man has a Recurring Deposit Account in a bank for $3\dfrac{1}{2}$ years. If the rate of interest is 12% per annum and the man gets ₹ 10,206 on maturity, find the value of monthly installments.

Let man deposits ₹ x per month.

So,

P = ₹ x, n = (3 × 12 + 6) = 42 months and r = 12%

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

$\therefore I = ₹ x \times \dfrac{42 \times 43}{2 \times 12} \times \dfrac{12}{100} \\[1em] = ₹ x \times \dfrac{21 \times 43}{100} \\[1em] = ₹ \dfrac{903x}{100}$

Maturity value = Sum deposited + Interest

$= ₹ x \times 42 + ₹\dfrac{903x}{100} \\[1em] = ₹\dfrac{4200x + 903x}{100} \\[1em] = ₹\dfrac{5103x}{100}$

Given, maturity value = ₹ 10206.

$\therefore \dfrac{5103x}{100} = 10206 \\[1em] \Rightarrow x = \dfrac{10206 \times 100}{5103} \\[1em] \Rightarrow x = \dfrac{1020600}{5103} = ₹ 200.$

Hence, the man paid ₹ 200 per month.