A man arranges to pay off a debt of Rs 3600 by 40 annual instalments which form an arithmetic series.

Let $S_{40}$ denote the total loan amount to be paid in 40 annual instalments.

$\therefore S_{40}=3600$

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, $S_{40}-S_{30}=\frac{1}{3} \times 3600$

$\Rightarrow 3600-S_{30}=1200$

$\Rightarrow S_{30}=2400$

$\Rightarrow \frac{30}{2}\{2 a+(30-1) d\}=2400$

$\Rightarrow 15\{2 a+29 d\}=2400$

$\Rightarrow 2 a+29 d=160 \quad \ldots .(2)$

On solving equations $(1)$ and $(2)$, we get:

d = 2 and a =51

Hence, the value of the first instalment is Rs 51