Mohit borrowed a certain sum at 5% per annum compound interest and cleared this loan by paying ₹ 12600 at the end of the first year and ₹ 17640 at the end of the second year. Find the sum borrowed.

If the letters have usual meanings, the formula for finding compound interest, compounded yearly for the given time is :

$1\dfrac{1}{2}$ years :

1. $P\Big[\Big(1 + \dfrac{r}{100}\Big)^{1\dfrac{1}{2}}- 1\Big]$

2. $P\Big(1 + \dfrac{r}{100}\Big)\Big(1 + \dfrac{r}{200}\Big)$ - P

3. $P\Big(1 + \dfrac{r}{200}\Big)^3$

4. $P\Big(1 + \dfrac{r}{100}\Big)^3\Big(1 + \dfrac{r}{200}\Big)- P$

If the letters have usual meanings, the formula for finding compound interest, compounded yearly for the given time is :

$2\dfrac{1}{2}$ years :

1. $P\Big(1 + \dfrac{r}{100}\Big)^{\dfrac{5}{2}} - P$

2. $P\Big(1 + \dfrac{r}{100}\Big)^{\dfrac{5}{2}}$

3. $P\Big(1 + \dfrac{r}{100}\Big)^2\Big(1 + \dfrac{r}{200}\Big)$

4. $P\Big(1 + \dfrac{r}{100}\Big)^2\Big(1 + \dfrac{r}{200}\Big)$ - P

When interest is compounded half-yearly then the formula for C.I. for the given time is :

1 year :

1. $P\Big(1 + \dfrac{r}{200}\Big)^2 - P$

2. $P\Big(1 + \dfrac{r}{100}\Big)^2 - P$

3. $P\Big(1 + \dfrac{r}{200}\Big)^2$

4. $P\Big(1 + \dfrac{r}{100}\Big)^2$