This can be model using the exponential growth formula:
[tex]\begin{gathered} y=a\cdot b^x \\ _{\text{ }}where\colon \\ b=1+r \end{gathered}[/tex]a = Initial amount = 821
r = Growth rate per time period = 2% = 0.02
x = time period
So:
[tex]y=821(1.02)^x[/tex]---
For x = 1/2 year = 6 months
[tex]\begin{gathered} y=821(1.02)^6 \\ y\approx925 \end{gathered}[/tex]-----
For x = 10 years = 120 months
[tex]\begin{gathered} y=821(1.02)^{120} \\ y\approx8838 \end{gathered}[/tex]-----------------------
For x = 100 years = 1200 months
[tex]\begin{gathered} y=821(1.02)^{1200} \\ y\approx1.7\times10^{13} \end{gathered}[/tex]