Question

A farmer borrows ₹ 1,000 and agrees to repay with a total interest of ₹ 140 in 12 installments. Each installment being less than the preceding installment by ₹ 10. What should be the amount of his first and last installment?

Answer 1

As each installment is less than the preceding installment by ₹ 10, the installments are in A.P.

S₁₂ = 1000 + 140 = 1140

n = 12, d = −10

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

`1140 = 12/2 [2a + (12 - 1)(-10)]`

`1140 = 6 [2a + (11)(-10)]`

`1140=6[2a-110]`

`1140/6=[2a-110]`

190 = [2a − 110]

2a = 190 + 110

2a = 300

a = `300/2`

a = 150

The first installment is ₹ 150.

The last installment (l) is the 12th term (T₁₂) of the AP:

T_n ​= a + (n − 1)d

T₁₂​ = 150 + (12 − 1)(−10)

T₁₂​ = 150 − 110

T₁₂​ = 40

The last installment is ₹ 40.