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