In general, the explicit formula of a geometric series is
[tex]a_n=a_{}r^{n-1}[/tex]
Where a and r are constants, and a_n is the n-th term.
In our case,
[tex]\begin{gathered} a_3=20 \\ \text{and} \\ a_3=ar^{3-1}=ar^2 \\ \Rightarrow ar^2=20 \end{gathered}[/tex]
On the other hand,
[tex]\begin{gathered} a_5=80 \\ \text{and} \\ a_5=ar^{5-1}=ar^4 \\ \Rightarrow ar^4=80 \end{gathered}[/tex]
Use the two equations to find a and r, as shown below
[tex]\begin{gathered} ar^2=20 \\ \Rightarrow r^2=\frac{20}{a} \\ \Rightarrow a(r^2)^2=80 \\ \Rightarrow a(\frac{20}{a})^2=80 \\ \Rightarrow\frac{400}{a}=80 \\ \Rightarrow a=5 \end{gathered}[/tex]
Finding r,
[tex]\begin{gathered} a=5 \\ \Rightarrow5r^2=20 \\ \Rightarrow r^2=4 \\ \Rightarrow r=\sqrt[]{4}=\pm2 \end{gathered}[/tex]
Thus, the explicit formula is
[tex]a_n=5(\pm2)^{n-1}[/tex]
Set n=1 and find a_1 as shown below,
[tex]\begin{gathered} n=1 \\ \Rightarrow a_1=5(\pm2)^{1-1}=5 \\ \Rightarrow a_1=5 \end{gathered}[/tex]
The answer is a_1=5
