Question

A deposit of N100 is made at the beginning of each month in an account that pays 6% interest, compounded monthly. The balance in the account at the end of 5 years is? 

Answer 1

A deposit of $100 is made at the beginning of each month in an account that pays 6% interest, compounded monthly. The balance A in the account at the end of 5 years is

$A=100(1 +\frac{0.06}{12})^1 +...+100(1 +\frac{0.06}{12})^{60}$

The balance A in account at the end of 5 years is,

$A=100(1 +\frac{0.06}{12})^1 + 100(1 + \frac{0.06}{12})^2+...+100(1 +\frac{0.06}{12})^{60}$

We can see that,

$a_1=100(1+\frac{0.06}{12}) \\ r=(1 +\frac{0.06}{12}) \\ n=60$

Now, we can apply the formula for the sum of a finite geometric sequence,

$A=S_n \\ =a(\frac{1-r^n}{1-r}) \\ =100(1+\frac{0.06}{12})(\frac{1-1.005^{60}}{1-1.005}) \\ = 7011.89$

Therefore the balance A in account at the end of 5 years is 7011.89