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