Solution:

As per the question:

There are 40 annual instalments that form an arithmetic series.

Let ‘$a$’ be the first instalment

$S_{40}=3600, n=40$

Using the formula,

$\begin{array}{l}
S_{n}=n / 2[2 a+(n-1) d] \\
3600=40 / 2[2 a+(40-1) d] \\
3600=20[2 a+39 d] \\
3600 / 20=2 a+39 d
\end{array}$
$2 a+39 d-180=0 \ldots \text { (i) }$

Given that,

The sum of first 30 terms is paid and one third of the debt is unpaid.
Therefore, paid amount $=2 / 3 \times 3600=$ ₹ 2400

$\mathrm{S}_{\mathrm{n}}=2400, \mathrm{n}=30$
Using the formula,
$\begin{array}{l}
S_{n}=n / 2[2 a+(n-1) d] \\
2400=30 / 2[2 a+(30-1) d] \\
2400=15[2 a+29 d] \\
2400 / 15=2 a+29 d \\
2 a+29 d-160=0 \ldots \text { (ii) }
\end{array}$

Let’s now solve eq.(i) and eq.(ii) by substitution method, we obtain
$\begin{array}{l}
2 a+39 d=180 \\
2 a=180-39 d \ldots \text { (iii) }
\end{array}$

On substituting the value of $2 \mathrm{a}$ in eq.(ii) we get,
$\begin{array}{l}
2 a+29 d-160=0 \\
180-39 d+29 d-160=0 \\
20-10 d=0 \\
10 d=20 \\
d=20 / 10 \\
=2
\end{array}$

Substituting the value of $d$ in eq.(iii)
$\begin{array}{l}
2 a=180-39 d \\
2 a=180-39(2) \\
2 a=180-78 \\
2 a=102 \\
a=102 / 2 \\
=51
\end{array}$

As a result, value of first installment ‘a’ is ₹ 51