Answer :

Given: A man borrows = Rs. 8000

Repay with total interest = Rs 1360

In 12 months, thus n = 12

Thus, S_{12} = 8000 + 1360 = 9360

Each installment being less than preceding one

Thus, d = – 40

We need to find “a”

Now, By using sum of n^{th} term of an A.P. we will find it’s sum

Where, n = no. of terms

a = first term

d = common difference

S_{n} = 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 n^{th} term of an A.P. we will find it’s sum

Where, n = no. of terms

S_{n} = sum of n terms

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

Let a = first term, t_{n} = last term

⇒ 9360 = 6 [ 1000 + t_{n}]

⇒ t_{n} = 1560 – 1000 = 560

Thus, last installment t_{n} = 560

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