A man arranges to pay off a debt of ₹36000 by 40 monthly instalments which form an arithmetic series.

Solution:

Let the value of the first instalment be ₹a.

Since the monthly instalments form an arithmetic series, so let us suppose the man increases the value of each instalment by ₹d every month.

∴ Common difference of the arithmetic series = ₹d

Amount paid in 30 instalments $=₹ 36,000-\frac{1}{3} \times ₹ 36,000=₹ 36,000-₹ 12,000=₹ 24,000$

Let Sn denote the total amount of money paid in the n instalments. Then,

$S_{30}=₹ 24000$

$\Rightarrow \frac{30}{2}[2 a+(30-1) d]=24000 \quad\left\{S_{n}=\frac{n}{2}[2 a+(n-1) d]\right\}$

$\Rightarrow 15(2 a+29 d)=24000$

$\Rightarrow 2 a+29 d=1600 \quad \ldots(1)$

Also,

$S_{40}=₹ 36,000$

$\Rightarrow \frac{40}{2}[2 a+(40-1) d]=36000$

$\Rightarrow 20(2 a+39 d)=36000$

$\Rightarrow 2 a+39 d=1800 \quad \ldots(2)$

Subtracting (1) from (2), we get

$(2 a+39 d)-(2 a+29 d)=1800-1600$

$\Rightarrow 10 d=200$

$\Rightarrow d=20$

Putting d = 20 in (1), we get

$2 a+29 \times 20=1600$

$\Rightarrow 2 a+580=1600$

$\Rightarrow 2 a=1600-580=1020$

$\Rightarrow a=510$

Thus, the value of the first instalment is ₹510.