Question

A man arranges to pay off a debt of ₹36000 by 40 monthly instalments which form an arithmetic series. When 30 of the instalments are paid, he dies leaving one-third of the debt unpaid. Find the value of the first instalment.

Answer 1

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 − $\frac{1}{3}$ × â‚¹36,000 = ₹36,000 − ₹12,000 = ₹24,000

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

S₃₀ = ₹24,000
$$\begin{gathered} \Rightarrow \frac{30}{2}\left[2a+\left(30-1\right)d\right]=24000\left\{S_n=\frac{n}{2}\left[2a+\left(n-1\right)d\right]\right\} \\ \Rightarrow 15\left(2a+29d\right)=24000 \\ \Rightarrow 2a+29d=1600.....\left(1\right) \end{gathered}$$

Also,

S₄₀ = ₹36,000
$$\begin{gathered} \Rightarrow \frac{40}{2}\left[2a+\left(40-1\right)d\right]=36000 \\ \Rightarrow 20\left(2a+39d\right)=36000 \\ \Rightarrow 2a+39d=1800.....\left(2\right) \end{gathered}$$

Subtracting (1) from (2), we get

$$\begin{gathered} \left(2a+39d\right)-\left(2a+29d\right)=1800-1600 \\ \Rightarrow 10d=200 \\ \Rightarrow d=20 \end{gathered}$$

Putting d = 20 in (1), we get

$$\begin{gathered} 2a+29\times 20=1600 \\ \Rightarrow 2a+580=1600 \\ \Rightarrow 2a=1600-580=1020 \\ \Rightarrow a=510 \end{gathered}$$

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