Answer
$a_1= -\frac{3}{2}, a_2= -\frac{3}{4}, a_3= -\frac{3}{8}, a_4= -\frac{3}{16}, a_5= -\frac{3}{32}, $ geometric.
Work Step by Step
Given $a_n=-3(\frac{1}{2})^n$, we have $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}, a_5=-3(\frac{1}{2})^5=-\frac{3}{32}, $ and we can identify the sequence as geometric.
