general geometric formula is
[tex]A_n=A_1\cdot r^{n-1}[/tex]
then we replace using A3
[tex]\begin{gathered} A_3=A_1\cdot r^{3-1} \\ \\ \frac{16}{3}=A_1\cdot r^2 \end{gathered}[/tex]
now replace using A5
[tex]\begin{gathered} A_5=A_1\cdot r^{5-1} \\ \\ \frac{64}{21}=A_1\cdot r^4 \end{gathered}[/tex]
now we have two equations and two unknow
[tex]\begin{gathered} \frac{16}{3}=A_1\cdot r^2 \\ \\ \frac{64}{21}=A_1\cdot r^4 \end{gathered}[/tex]
we can solve A1 or r from any equation and replace on the other
I will solve A1 from the first equation
[tex]A_1=\frac{\frac{16}{3}}{r^2}[/tex]
and replace on the second to solve r
[tex]\begin{gathered} \frac{64}{21}=(\frac{\frac{16}{3}}{r^2})\cdot r^4 \\ \\ \frac{64}{21}=\frac{16}{3}\cdot r^2 \\ \\ r^2=\frac{64\times3}{21\times16} \\ \\ r^2=\frac{192}{336}=\frac{4}{7} \\ \\ r=\frac{2\sqrt[]{7}}{7} \end{gathered}[/tex]
now replace r on the other equation to find A1
[tex]\begin{gathered} \frac{16}{3}=A_1\cdot(\frac{2\sqrt[]{7}}{7})^2 \\ \\ \frac{16}{3}=A_1\cdot\frac{4}{7} \\ \\ A_1=\frac{16\times7}{4\times3} \\ \\ A_1=\frac{28}{3} \end{gathered}[/tex]
now we have the two unknows A1 and r then replace on the general geometric equation
[tex]A_n=\frac{28}{3}\cdot(\frac{2\sqrt[]{7}}{7})^{n-1}[/tex]
