Question

A man borrows Rs 8000 and agrees to repay with a total interest of Rs 1360 in 12 monthly installments. Each installment being less than the preceding one by Rs 40. Find the amount of the first and last installments.

Answer 1

Money he borrows = Rs. 8000

Total interest = Rs. 1360 

Total money he will pay after 12 months = Rs (8000 + 1360) = Rs 9360
n = 12 and S₁₂ = 9360

Each installment being less than the preceding one by Rs 40.
d = −40
Now,

S_n = `n/2 [2a + (n − 1)d]`

∴ S₁₂ = `12/2 [2a + (12 − 1)(− 40)]`

∴ 9360 = 6[2a + 11 × (− 40)]

∴ 9360 = 6(2a − 440)

∴ `9360/6 = 2a − 440`

∴ 1560 = 2a − 440

∴ 1560 + 440 = 2a

∴ 2000 = 2a

∴ a = `2000/2`

∴ a = 1000

t_n = a + (n − 1)d

∴ t₁₂ = 1000 + (12 − 1)(− 40)

∴ t₁₂ = 1000 + 11(− 40)

∴ t₁₂ =  1000 − 440

∴ t₁₂ = 560

∴ Amount of the first instalment is Rs. 1000 and that of the last instalment is Rs. 560.