Question

A man arranges to pay-off a debt of $₹ 3600$ by $40$ annual instalments, which are in A.P. . When $30$ of the instalments are paid, he dies leaving one-third of the debt unpaid. The $8^{\text {th }}$ instalment is

• A. ₹ 60
• B. ₹ 70
• C. ₹ 65
• D. ₹ 75

Answer 1

Answer: (C)

Here, total debt, $S=₹ 3600$
and total instalments, $n=40$
Let $a$ and $d$ be the first instalment and increment in instalment.
Thus, we get an A.P. here.
By using sum of $n$ terms, we get
$ 3600 =\frac{40}{2}[2 a+(40-1) d]$
$\Rightarrow 180 =2 a+39 d ....$(i)
Now, after 30 instalments, one-third of the debt is unpaid.
i.e., $\frac{3600}{3}=₹ 1200$ is unpaid and then paid money
$=3600-1200=₹ 2400$
So, again by using sum of $n$ terms, we get
$S_{30}=2400 =\frac{30}{2}[2 a+(30-1) d]$
$\Rightarrow 160 =2 a+29 d$...(ii)
On solving Eqs. (i) and (ii), we get
$a=51, d=2$
We know that,
$n^{\text {th }} \text { term } T_n =a+(n-1) d$
$\therefore 8^{\text {th }} $ instalment, $ T_8 =a+(8-1) d$
$=51+7 \times 2=51+14=₹ 65$
Hence, the $8^{\text {th }}$ instalment is $₹ 65$.