Question

An amount $Rs.500$ is deposited in a bank which pays annual interest at the rate of $10$% compounded annually. What will be the value of this deposit at the end of ${10}^{th}$ year?

Answer 1

The principal is Rs. $500$.

So, the interest for this principal for one year is $500\left(\frac{10}{100}\right)=50$.
Thus, the principal for the $2^{nd}$ year $=$ Principal for $1^{st}$ year $+$ Interest
$=500+500\left(\frac{10}{100}\right)=500(1+\frac{10}{100})$
Now, the interest for the second year $=\left(500(1+\frac{10}{100})\right)(1+\frac{10}{100})$.
So, the principal for the third year $=500(1+\frac{10}{100})+500(1+\frac{10}{100})\frac{10}{100}$
$=500{(1+\frac{10}{100})}^2$
Continuing in this way we see that the principal for the $n^{th}$ year $=500{(1+\frac{10}{100})}^{n-1}$
The amount at the end of $(n-1{)}^{th}year=$ Principal for the $n^{th}$ year.
Thus, the amount in the account at the end of $n^{th}$ year.
$=500{(1+\frac{10}{100})}^{n-1}=500{(1+\frac{10}{100})}^{n-1}\left(\frac{10}{100}\right)=500{\left(\frac{11}{10}\right)}^n$.
The amount in the account at the end of ${10}^{th}$ year
$=$ Rs. $500{(1+\frac{10}{100})}^{10}=$ Rs.$500{\left(\frac{11}{10}\right)}^{10}$.