Answer
$a_1=-\dfrac{3}{2}$
$a_2=-\dfrac{3}{4}$
$a_3=-\dfrac{3}{8}$
$a_4=-\dfrac{3}{16}$
The sequence is geometric with a common ratio of $\dfrac{1}{2}$.
Work Step by Step
We need to substitute $1, 2, 3,$ and $4$ for $n$ into the given equation to find the first four terms.
$a_1=-3(\dfrac{1}{2})^1 = -3(\dfrac{1}{2})=-\dfrac{3}{2}$
$a_2=-3(\dfrac{1}{2})^2 = -3(\dfrac{1}{4})=-\dfrac{3}{4}$
$a_3=-3(\dfrac{1}{2})^3 = -3(\dfrac{1}{8})=-\dfrac{3}{8}$
$a_4=-3(\dfrac{1}{2})^4 = -3(\dfrac{1}{16})=-\dfrac{3}{16}$
We see that the the next term is equal to $\dfrac{1}{2}$ times the current term and this implies that the sequence is geometric with a common ratio of $\dfrac{1}{2}$.
