Answer
The sequence is geometric with a common ratio of $\frac{1}{2}$.
The first four terms are:
$a_1=-\frac{3}{2}$
$a_2=-\frac{3}{4}$
$a_3=-\frac{3}{8}$
$a_4=-\frac{3}{16}$
Work Step by Step
Substitute $1, 2, 3,$ and $4$ for $n$ into the given formula:
$a_1=-3(\frac{1}{2})^1 = -3(\frac{1}{2})=-\frac{3}{2}$
$a_2=-3(\frac{1}{2})^2 = -3(\frac{1}{4})=-\frac{3}{4}$
$a_3=-3(\frac{1}{2})^3 = -3(\frac{1}{8})=-\frac{3}{8}$
$a_4=-3(\frac{1}{2})^4 = -3(\frac{1}{16})=-\frac{3}{16}$
Notice the the next term is equal to $\frac{1}{2}$ times the current term.
This means that a common ratio of $\frac{1}{2}$ exists, and the sequence is geometric.