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