A man arranges to pay off a debt of Rs 3600 by 40 annual 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.

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Solution

Let $S_{40}$ denote the total loan amount to be paid in 40 annual instalments.
∴ $S_{40}$ = 3600
Let Rs a be the value of the first instalment and Rs d be the common difference.
We know:

$S_{n} = \frac{n}{2} \{2 a + (n - 1) d\}$

$\Rightarrow \frac{40}{2} \{2 a + (40 - 1) d\} = 3600 \Rightarrow 20 \{2 a + 39 d\} = 3600 \Rightarrow 2 a + 39 d = 180 . . . . . . (1)$

$Also, S_{40} - S_{30} = \frac{1}{3} \times 3600 \Rightarrow 3600 - S_{30} = 1200 \Rightarrow S_{30} = 2400 \Rightarrow \frac{30}{2} \{2 a + (30 - 1) d\} = 2400 \Rightarrow 15 \{2 a + 29 d\} = 2400 \Rightarrow 2 a + 29 d = 160 . . . . . (2)$

On solving equations $(1) and (2)$ , we get:
d = 2 and a =51

Hence, the value of the first instalment is Rs 51

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