Answer
In order for a sequence to be geometric, the quotient of all consecutive terms must be constant.
Hence here: $\frac{a_{n+1}}{a_n}=\dfrac{(\frac{3^{(n+1)}}{9})}{(\frac{3^{n}}{9})}=3$, thus it is a geometric sequence.
$a_1=\frac{1}{3}$
$a_2=1$
$a_3=3$
$a_4=9$
Work Step by Step
In order for a sequence to be geometric, the quotient of all consecutive terms must be constant.
Hence here: $\frac{a_{n+1}}{a_n}=\dfrac{(\frac{3^{(n+1)}}{9})}{(\frac{3^{n}}{9})}=3$, thus it is a geometric sequence.
$a_1=\frac{3^{1}}{9}=\frac{3}{9}=\frac{1}{3}$
$a_2=\frac{3^{2}}{9}=\frac{9}{9}=1$
$a_3=\frac{3^{3}}{9}=\frac{27}{9}=3$
$a_4=\frac{3^{4}}{9}=\frac{81}{9}=9$
