# A man arranges too pay off a debt of Rs 3600 by 40 annual installments which form an 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.

Given:

Total amount of debt to be oaid in 40 installments = Rs 3600

After 30 installments, one-third of his debt is unpaid. This means that two-third of his payment is done.

Therefore, the amount he paid in 30 installments = 2/3(3600) = 2(1200) = 2400

Let a be the first installment and d be the common difference between them.

We know from the formula of AP that,

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

Substituting the values of a and d,

S30 = 30/2[2a+(30-1)d]

2400=15[2a+29d]

a=160-29d/2 eq.1

Similarly, a and d are calculated for 40 installments

S40 = 40/2[2a+(40-1)d]

3600=20(2a+39d)

a=180-39d/2 eq.2

Subtracting eq.1 and eq.2

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

0=20-10d

10d=20

d=2

Substituting the value of d,

a=160-29(2)/2

a=51

Therefore, Rs.51 was paid in the first installment.