smc

Let the first term be $a$ and the common ratio be $r$. Therefore, $$ar=2(1)\text{and}a{r}^3=6(2)$$ Dividing $(2)$ by $(1)$ eliminates the $a$, yielding $r^2=3$, so $r=\pm \sqrt{3}$. Now, since $ar=2$, $a=\frac{2}{r}$, so $a=\frac{2}{\pm \sqrt{3}}=\pm \frac{2\sqrt{3}}{3}$. We therefore see that $\boxed{-\frac{2\sqrt{3}}{3}}$ is a possible first term.