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