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