A man arranges to pay off a debt of ₹36000 by 40 annual instalments which

Solution:

Given: -

Total debt = Rs. 36000

A man pays this debt in 40 annual instalments that forms an A.P.

After annual instalments, that man dies leaving one - third of the debt unpaid.

So,

Within 30 instalments he pays two - thirds of his debt.

Sum of $n$ terms in an Arithmetic Progression $=\frac{n}{2}[2 \times a+(n-1) \times d]$

He has to pay 36000 in 40 annual instalments,

$36000=\frac{40}{2}[2 \times a+(40-1) \times d] \rightarrow(1)$

Where,

$a=$ amount paid in the first instalment,

$d=$ difference between two Consecutive instalments.

He paid two - a third of the debt in 30 instalments,

$\frac{2}{3}(36000)=\frac{30}{2}[2 \times a+(30-1) \times d] \rightarrow(2)$

From equations (1) & (2) we get,

a = 510 & d = 20

∴The value of the first instalment is Rs.510.