Answer
a) $\$11616.77$.
b) $\$1616.77$.
Work Step by Step
(a)
In the given case, $P=\$2000$ and $r=0.075$ for $t=5\text{ years}$.
It is compounded annually, so $n=1$.
Thus:
$\begin{align}
& A=\frac{2000\left[ {{\left( 1+\frac{0.075}{1} \right)}^{5}}-1 \right]}{\left( \frac{0.075}{1} \right)} \\
& =\frac{2000\left[ {{\left( 1.075 \right)}^{5}}-1 \right]}{\left( 0.075 \right)} \\
& =\frac{2000\left[ \left( 1.4356 \right)-1 \right]}{\left( 0.075 \right)} \\
& =11616.77
\end{align}$
(b)
Here $P=\$2000$ and $r=0.075$ for $t=5\,years$ and it is compounded annually, so $n=1$.
Thus:
$\begin{align}
& A=\frac{2000\left[ {{\left( 1+\frac{0.075}{1} \right)}^{5}}-1 \right]}{\left( \frac{0.075}{1} \right)} \\
& =\frac{2000\left[ {{\left( 1.075 \right)}^{5}}-1 \right]}{\left( 0.075 \right)} \\
& =\frac{2000\left[ \left( 1.4356 \right)-1 \right]}{\left( 0.075 \right)} \\
& =11616.77
\end{align}$
Now we can find the interest as below:
$Interest=A-P$ and total deposit = $P\times n\times t$.
So;
$\begin{align}
& Interest=11616.77-2000\times 1\times 5 \\
& =11616.77-10000 \\
& =1616.77
\end{align}$
