A man arranges to pay off a debt of Rs. 36000 by 40 monthly 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 the first installment = Rs. x

Since the instalments form an arithmetic series, therefore let the common difference = d


Now, amount paid in 30 installments = two - third of the amount = (2/3) × (36000) = Rs. 24000


∴ Total amount paid by the man in 30 installments = 24000


Let Sn be that amount paid in 30 installments.


∴ S30 = 24000


⇒ (30/2) × [2x + (30 - 1)d] = 24000


⇒ 15 × [2x + 29d] = 24000


⇒ 2x + 29d = 1600 …………..(1)


Now, Total sum of the amount = 36000


∴ S40 = 36000


⇒ (40/2) × [2x + (40 - 1)d] = 36000


⇒ 20 × [2x + 39d] = 36000


⇒ 2x + 39d = 1800 ………..(2)


Subtracting equation (1) from equation (2), we get:


10d = 200


⇒ d = 20


∴ from equation (1), we get


x = 1/2(1600 - 29d)


= 1/2 (1600 - 580)


= 1/2 (1020)


= 510


Therefore the amount of first installment = Rs. 510


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