Answer
In order for a sequence to be geometric, the quotient of all consecutive terms must be constant.
Hence here: $\frac{a_{n+1}}{a_n}=\dfrac{(\frac{3^{((n-1)+1)}}{2^{n+1}})}{(\frac{3^{(n-1)}}{2^n})}=\frac{3}{2}$, thus it is a geometric sequence.
$a_1=\frac{1}{2}$
$a_2=\frac{3}{4}$
$a_3=\frac{9}{8}$
$a_4=\frac{27}{16}$
Work Step by Step
In order for a sequence to be geometric, the quotient of all consecutive terms must be constant.
Hence here: $\frac{a_{n+1}}{a_n}=\dfrac{(\frac{3^{((n-1)+1)}}{2^{n+1}})}{(\frac{3^{(n-1)}}{2^n})}=\frac{3}{2}$, thus it is a geometric sequence.
$a_1=\frac{3^{(1-1)}}{2^1}=\frac{1}{2}$
$a_2=\frac{3^{(2-1)}}{2^2}=\frac{3}{4}$
$a_3=\frac{3^{(3-1)}}{2^3}=\frac{9}{8}$
$a_4=\frac{3^{(4-1)}}{2^4}=\frac{27}{16}$
