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

1.	A man arranges to pay off a debt of ₹36000 by 40 annual instalments which form an AP. When 30 of the instalments are paid, he dies, leaving one - third of the debt unpaid. Find the value of the first instalment.
Answer» 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 = n/2[2 x a+(n-1) x d] He has to pay 36000 in 40 annual instalments, 36000 = 40/2[2 x a+(40 -1) x d] → (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, 2/3(36000) = 30/2[2 x a+(30 -1) x d] → (2) From equations (1) & (2) we get, a = 510 & d = 20 ∴The value of the first instalment is Rs.510.

A man arranges to pay off a debt of ₹36000 by 40 annual instalments which form an AP. When 30 of the instalments are paid, he dies, leaving one - third of the debt unpaid. Find the value of the first instalment.

Answer»

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 = n/2[2 x a+(n-1) x d]

He has to pay 36000 in 40 annual instalments,

36000 = 40/2[2 x a+(40 -1) x d] → (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,

2/3(36000) = 30/2[2 x a+(30 -1) x d] → (2)

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

a = 510 & d = 20

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