Question

A man arranges to pay off a debt of Rs 3600 by 40 annual instalments which form an arithmetic series. When 30 of the instalments are paid, he dies leaving one-third of all debt unpaid, finds the value of the first instalment.

Answer 1

In the given problem,

The total amount of debt to be paid in 40 installments = Rs 3600

After 30 installments one−third of his debt is left unpaid. This means that he paid two third of the debt in 30 installments. So,

The amount he paid in 30 installments  = `2/3 (3600)`

= 2(1200)

= 2400

Let us take the first installment as a and common difference as d.

So, using the formula for the sum of n terms of an A.P,

`S_n = n/2[2a + (n -1)d]`

Let us find a and d, for 30 installments.

`S_30 = 30/2[2a + (30 - 1)d]`

2400 = 15 [2a + (29)d]

`2400/15 = 2a + 29d`

160 = 2a + 29d

`a = (160 - 29d)/2`  .....(1)

Similarly, we find a and d for 40 installments.

`S_40 = 40/2[2a+ (40 - 1)d]`

3600 = 20[2a + (39)d]

`3600/20 = 2a + 39d`

`a = (180 - 39d)/2` .....(2)

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

`a- a = ((180 - 39d)/2)- ((160 - 29d)/2)`

`0 = (180 - 39d - 160 + 29d)/2`

0 = 20 - 10d

Further solving for d

10d = 20

`d = 20/10`

d = 2

Substituting the value of d in (1), we get.

`a = (160 - 29(2))/2`

`= (160 - 58)/2`

`= 102/2`

= 51

Therefore, the first installment is Rs 51