Answer
Geometric sequence.
Work Step by Step
In order for a sequence to be geometric, the quotient of all consecutive terms must be constant.
Here: $\frac{a_{n+1}}{a_n}=\frac{(-\frac{1}{3})^{n+1}}{(-\frac{1}{3})^{n}}=\frac{(-\frac{1}{3})(-\frac{1}{3})^{n}}{(-\frac{1}{3})^{n}}=-\frac{1}{3}$
Thus we can see that the quotient is constant, thus it is a geometric sequence.