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# 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.

**Given:**

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.

**To do:**

We have to find the value of the first instalment.

**Solution:**

Total debt \( =Rs. 3600 \)

Number of instalments (terms) \( =40 \)

\( \therefore \mathrm{S}_{40}=3600 \)

\( S_{40}-S_{30}=\frac{1}{3} \text { of debt }=\frac{1}{3} \times 3600 \)

\( =1200 \)

\( \therefore S_{30}=3600 \times \frac{2}{3}=Rs. 2400 \)

Let \( a \) be the first instalment and \( d \) the common difference.

We know that,

\( \mathrm{S}_{n}=\frac{n}{2}[2 a-(n-1) d] \)

\( \mathrm{S}_{40}=\frac{40}{2}[2 a-(40-1) d] \)

\( 3600= 20(2 a-39 d) \)

\( 180= 2 a-39 d \).......(i)

\( \mathrm{S}_{30}=\frac{30}{2}[2 a-(30-1) d] \)

\( 2400 = 15(2 a-29 d) \)

\( 160 = 2 a-29 d \).......(ii)

Subtracting (ii) from (i) \( 10 d=20 \)

\( \Rightarrow d=\frac{20}{10}=2 \)

\( 2 a+39 d=180 \)

\( \Rightarrow 2 a+39 \times 2=180 \)

\( \Rightarrow 2 a+78=180 \)

\( \Rightarrow 2 a=180-78=102 \)

\( \therefore a=\frac{102}{2}=51 \)

**Hence, the value of the first instalment is \( Rs.\ 51 \).**