Answer
a.
$A_{1}=\$ 2004$,
$A_{2}=\$ 2008.01$,
$A_{3}=\$ 2012.02$,
$A_{4}=\$ 2016.05$,
$A_{5}=\$ 2020.08$,
$A_{6}=\$ 2024.12$
b.
$A_{36}=\$ 2149.16$.
Work Step by Step
a.
Using our calculator, $A_{n}=2000(1+\displaystyle \frac{0.024}{12})^{n},$
$A_{1}=\$ 2004$,
$A_{2}=\$ 2008.01$,
$A_{3}=\$ 2012.02$,
$A_{4}=\$ 2016.05$,
$A_{5}=\$ 2020.08$,
$A_{6}=\$ 2024.12$
b.
3 years = 36 months
n=36.
$A_{36}=\$ 2149.16$.
