# A man repays a loan of 6500 by paying 610 in the first month and decreases the payment by 30 each month find the time that will be taken

1
by gill1

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by gill1

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There is no mention of interest here. Hence we just add the repayments.

Loan P = 6, 500

repayments = 610 + 580 + 550 + 520 + .....n terms

common difference = 30 . This is arithmetic progression.

Nth term = 610 - (n-1) 30

sum = [610 + 610 - (n-1) 30] * N / 2 = 6, 500

N [1, 220 - 30 (N-1) ] = 13, 000

125 N - 3 N² = 1, 300

3 N² - 125 N + 1,300

N = [ 125 +- √[ 125² - 15600] ] /6 =

N = 20

In 20 months he is able to repay the loan.

Loan P = 6, 500

repayments = 610 + 580 + 550 + 520 + .....n terms

common difference = 30 . This is arithmetic progression.

Nth term = 610 - (n-1) 30

sum = [610 + 610 - (n-1) 30] * N / 2 = 6, 500

N [1, 220 - 30 (N-1) ] = 13, 000

125 N - 3 N² = 1, 300

3 N² - 125 N + 1,300

N = [ 125 +- √[ 125² - 15600] ] /6 =

N = 20

In 20 months he is able to repay the loan.