Question

After 10 years, the investment compounded periodically will be worth XXXX more than the investment compounded annually.

Answer 1

Given:

$P=\text{ \ 3000 ; interest rate(r)=}9\text{ \%= 0.09 ; t=10 years}$

Amount compounded annually, n=1

$A=P(1+\frac{r}{n})^{nt}$$\begin{gathered} A=3000(1+0.09)^{10} \\ A=3000(1.09)^{10} \\ A=7102.09 \\ A=\text{ \ 7102.09} \end{gathered}$

Amount compounded 2 periods in a year, n=2

$\begin{gathered} A=3000(1+\frac{0.09}{2})^{20} \\ A=3000(\frac{2.09}{2})^{20} \\ A=\text{ \ 7235.14} \end{gathered}$$\begin{gathered} \text{Difference amount=7235.14-7102.09} \\ \text{Difference amount= \ 133.05} \end{gathered}$