# 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 $Rs.\ 51$.

Updated on: 10-Oct-2022

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