Answer :

Given: A man borrows = Rs. 8000

Repay with total interest = Rs 1360


In 12 months, thus n = 12


Thus, S12 = 8000 + 1360 = 9360


Each installment being less than preceding one


Thus, d = – 40


We need to find “a”


Now, By using sum of nth term of an A.P. we will find it’s sum



Where, n = no. of terms


a = first term


d = common difference


Sn = sum of n terms


Thus, on substituting the given value in formula we get,



9360 = 6 [ 2a – 11 × 40]



1560 = 2a – 440


1560 + 440 = 2a


2a = 2000



Thus, first installment a = Rs. 1000


Now, By using sum of nth term of an A.P. we will find it’s sum



Where, n = no. of terms


Sn = sum of n terms


Thus, on substituting the given value in formula we get,


Let a = first term, tn = last term



9360 = 6 [ 1000 + tn]



tn = 1560 – 1000 = 560


Thus, last installment tn = 560


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