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Home » 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.
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.
RS aggarwal class 11 Maths Arithmetic progressions
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.

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 = 

n2[2a+(n–1)d]

He has to pay 36000 in 40 annual instalments,

36000=402[2×a+(40–1)×d]

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,

23(36000)=302[2a+(30–1)d]

From equations (1) & (2) we get,
a = 510 & d = 20
∴The value of the first instalment is Rs.510.

 

 

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