Q) Rohan repays his total loan of Rs.1,18,000 by paying every month starting with the first installment of Rs. 1,000. If he increases the installment by Rs. 100 every month, what amount will be paid by him in the 30th installment? What amount of loan has he paid after 30th installment?

Ans:

We can see that the loan repayment forms an A. P., where:

a = 1,000, d = 100, Sum of AP = 1,18,000

Let’s start from value of the 30th term (i.e. amount paid in this installment).

We know that nth term of an A.P.  =  a + (n-1) d

Therefore, T30

= 1000 + (30 – 1) x 100 = 3900

Therefore, the amount paid in 30th installment is Rs. 3,900.

Next, sum of n terms of an A.P.  Sn = \frac{n}{2} (2a + (n-1) d)

Therefore, S30 = \frac{30}{2} (2 x 1000 + (30 – 1) x 100)

= 15 (2000 + 2900) = 73500

Therefore the loan paid after 30th installment is Rs. 73,500.

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