Question

A man arranges to pay off debt of ₹36000 by 40 monthly instalments which form an arithmetic series. When 30 of the installments are paid, he dies leaving on-third of the debt
unpaid. Find the value of the first instalment.

Answer 1

Let the value of the first installment be ₹ a.
Since the monthly installments form an arithmetic series, so let us suppose the man increases the value of each installment by ₹ d every month.
∴ Common difference of the arithmetic series = ₹d
`"Amount paid in 30 installments " = RS  36,000 - 1/3 xx Rs 36,000=Rs 36000 - Rs 12,000=Rs 24,000.`

Let S_n denote the total amount of money paid in the n installments. Then, 

`S_30 = 24,000`

`⇒30/2 [2a +(30-1) d] = 24000                  {S_n = n/2 [2a + (n-1) d]}`

⇒ 15 (2a + 29d )= 24000

⇒ 2a +29d =1600                       .............(1)

Also,

S_{40  =} Rs 36,000

`⇒ 40/2 [2a +(40-1) d] = 36000`

⇒20 (2a +39d)=36000

⇒2a +39d=1800                 ................(2)

Subtracting (1) from (2), we get

(2a +39d) -(2a+29d) = 1800-1600

⇒ 10d = 200

⇒ d=20

Putting d = 20 in (1), we get

2a +29 × 20 = 1600

⇒ 2a + 580 = 1600

⇒ 2a = 1600 -580=1020

⇒ a=510

Thus, the value of the first installment is ₹510.