Step 1
State the formula for the future value.
[tex]FV=\text{PMT(}\frac{(1+i)^n-1}{i})[/tex]where;
[tex]\begin{gathered} FV=10,000 \\ i=0.02 \\ \text{PMT}=800 \\ n=\text{?} \end{gathered}[/tex]Step 2
Plug in the given values
[tex]\begin{gathered} 10000=800(\frac{(1+0.02)^n-1}{0.02}) \\ \frac{800\left(\left(1+0.02\right)^n-1\right)}{0.02}(0.02)=10000(0.02) \end{gathered}[/tex][tex]\begin{gathered} 800\mleft(\mleft(1+0.02\mright)^n-1\mright)=200 \\ (1+0.02)^n-1=\frac{200}{800} \\ (1+0.02)^n-1=0.25 \\ (1+0.02)^n=0.25+1 \\ (1+0.02)^n=1.25 \\ \ln (1+0.02)^n=\ln (1.25) \\ n\ln (1+0.02)=\ln (1.25) \\ n=\frac{\ln 1.25}{\ln (1.02)} \\ n=11.26838111 \\ n\approx11\text{ to the nearest integer} \end{gathered}[/tex]Hence, approximately to the nearest integer, n=11
