SOLUTION:
Given;
[tex]P=5000,r=2\text{ \%}=0.02[/tex](a) At the end of the first year;
[tex]\begin{gathered} A=P(1+\frac{r}{n})^{nt} \\ \\ A=5000(1+0.02)^1 \\ \\ A=5100 \end{gathered}[/tex]ANSWER: $5,100
(b) At the end of two years;
[tex]\begin{gathered} A=5000(1+0.02)^2 \\ \\ A=5202 \end{gathered}[/tex]ANSWER: $5,202