ANSWER
[tex]6[/tex]
EXPLANATION
We want to find the first term of the sequence.
The general equation for the nth term a geometric sequence is written as:
[tex]a_n=ar^{n-1}[/tex]
where a = first term; r = common ratio
Let us use this to write the equations for the third term and the fifth term.
For the third term, n = 3:
[tex]\begin{gathered} a_3=ar^2 \\ \Rightarrow54=ar^2 \end{gathered}[/tex]
For the fifth term, n = 5:
[tex]\begin{gathered} a_5=ar^4 \\ \Rightarrow486=ar^4 \end{gathered}[/tex]
Let us make a the subject of both formula:
[tex]\begin{gathered} 54=ar^2_{} \\ \Rightarrow a=\frac{54}{r^2} \end{gathered}[/tex]
and:
[tex]\begin{gathered} 486_{}=ar^4 \\ a=\frac{486}{r^4} \end{gathered}[/tex]
Now, equate both equations above and solve for r:
[tex]\begin{gathered} \frac{54}{r^2}=\frac{486}{r^4} \\ \Rightarrow\frac{r^4}{r^2}=\frac{486}{54} \\ \Rightarrow r^{4-2}=9 \\ \Rightarrow r^2=9 \\ \Rightarrow r=\sqrt[]{9} \\ r=3 \end{gathered}[/tex]
Now that we have the common ratio, we can solve for a using the first equation for a:
[tex]\begin{gathered} a=\frac{54}{r^2} \\ \Rightarrow a=\frac{54}{3^2}=\frac{54}{9} \\ a=6 \end{gathered}[/tex]
That is the first term.
