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