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

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 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