The answer is NO. This conclusion can be shown with proof by contradiction. Suppose it is possible. Let three prime numbers $p_1 < p_2 < p_3$ be three terms of a geometric sequence whose common ratio is $r$. Then, there exist two positive integers $m$ and $n$ such that $$p_2=p_1\cdot r^m\quad\text{and}\quad p_3=p_2\cdot r^n$$ Solving $r$ using these two equations, respectively, yields: $$r=p_1^{-\frac{1}{m}}\cdot p_2^{\frac{1}{m}}=p_2^{-\frac{1}{n}}\cdot p_3^{\frac{1}{n}}\implies p_2^{m+n}=p_1^{n}\cdot p_3^{m}$$ This means $p_2$ divides $p_3$ which is impossible.