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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.
Solution
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,
Sn = `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
tn = 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.
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