## Precalculus: Concepts Through Functions, A Unit Circle Approach to Trigonometry (3rd Edition)

The sequence is geometric with a common ratio of $\dfrac{3}{2}$.
We need to substitute $1, 2, 3,$ and $4$ for $n$ into the given equation to find the first four terms. $t_1=\dfrac{3^{1-1}}{2^1}=\dfrac{1}{2}$ $t_2=\dfrac{3^{2-1}}{2^2}=\dfrac{3}{4}$ $t_3=\dfrac{3^{3-1}}{2^3}=\dfrac{9}{8}$ $t_4=\dfrac{3^{4-1}}{2^4}=\dfrac{27}{16}$ We can see that the common ratios are the same; thus the sequence is geometric with a common ratio of $\dfrac{3}{2}$.