Answer:
[tex]\begin{gathered} a_n=2(\sqrt{3})^{n-1} \\ \\ a_9=162 \end{gathered}[/tex]
Step-by-step explanation:
Notice that the formula:
[tex]a_n=2(\sqrt{3})^{n-1}[/tex]
Has the general looking form of a series, and also stands for the a1 given, since:
[tex]a_1=2[/tex]
This way, we'll have that:
[tex]\begin{gathered} a_9=2(\sqrt{3})^{9-1} \\ \\ \Rightarrow a_9=162 \end{gathered}[/tex]
