# Simple and Compound Interests

Completion of squares Ahoy!

## SICI - Instalments

Q.15: Krishna borrows Rs. 45K from a bank at 10% compound interet. He repays it in three annual installments that are in arithmetic progression. He ends up paying 54K totally. How much did he pay in year 1?
1. Rs. 16,500
2. Rs. 19,500
3. Rs. 21,000
4. Rs. 18,000

Choice B. Rs. 19,500

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## Detailed Solution

Let the repayments be Rs "a – d", Rs "a" and Rs. "a + d"

a – d + a + a + d = 54000
3a = 54000
a = 18000

The payment at the end of year 2 is Rs. 18,000.
Borrowed amount = Rs. 45,000

Amount outstanding at the end of Year 1 = (45000 × 1.1) – (18000 – d)
= 31500 + d

Amount outstanding at the end of Year 2 = ((31500 + d) × 1.1) – 18000
= 34650 + 1.1d – 18000 = 16650 + 1.1d

Amount outstanding at the end of Year 3 = ((16650 + 1.1d) × 1.1) = 18000 + d
18315 + 1.21d = 18000 + d
0.21d = – 315
d = –1500

The payments are Rs. 19500, Rs. 18000 and Rs. 16500

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