Question

Aman arrange stop ay off a debt of Rs.3600 by 40 annual installments which for man arithmetic series. When 30 of the installments are paid, he dies leaving one - third of the debt unpaid, find the value of the first installment.

Answer 1

Given,

Total amount of payable in 40 annual installments, S₄₀ = 3600 

After 30 installment he died and leaving \(\frac{1}{3}\) of the debt unpaid 

Which means, total in 30 installment, 

S₃₀= \(\frac{2}{3}\) of the debt 

S₃₀= \(\frac{2}{3} \times\)3600 = 2400

We know sum of n terms, S_n = \(\frac{n}{2}\)[2a + (n – 1) d]

For 30 installments, S₃₀ = \(\frac{30}{2}\)[2a + (30 – 1) d]

2400 = 15 [2a + 29d] 

160 = 2a + 29d 

2a = 160 – 29d

a = \(\frac{160 - 29d}{2}\)..... (i)

For 40 installments, S₄₀ = \(\frac{n}{2}\)[2a + (n – 1) d]

3600 = \(\frac{40}{2}\)[2a + (40 – 1) d] 

180 = 2a + 39d 

2a = 180 – 39d

a = \(\frac{180-39d}{2}\) .... (ii)

From (i) and (ii), we get

\(\frac{160-29d}{2} = \frac{180-39d}{2}\)

160 – 29d = 180 – 39d 

39d – 29d = 180 – 160 

10d = 20 

d = 2

Putting the value of d in (i), we get

a = \(\frac{160 - 29(2)}{2}\)

\(\frac{160 - 58}{2} = \frac{102}{2} = 51\)

Hence, value of first installment is 51