A man arranges to pay off a debt of ₹ 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 instalment.

Given: Total debt is Rs. 3600 and total number of installments are 40 and in 30 installments, he has paid two-third of the debt and dies leaving one-third of amount


Let first instalment be “a”


Let the common difference of the instalments be “d”


Here, S40 = 3600 and n = 40


Formula used:



where a is first term, d is common difference and n is number of terms in an A.P.


Therefore,





2a+39d = 180 ………(1)


Since 30 installments are paid and one third of the debt unpaid



So,


S30 = 2400 and n = 30







2a+29d = 160 ………(2)


Solving (1) and(2) by substitution method,


2a + 39d = 180


2a = 180 – 39d ……(3)


Put value of 2a from (3) in (2):


2a + 29d = 160


180 – 39d + 29d = 160


–10d = 160 – 180


–10d = –20



d = 2


Put this value of d in (3):


2a = 180 – 39d


2a = 180 – 39(2)


2a = 180 – 78


2a = 102



a = 51


Hence, value of first installment = a = Rs. 51


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