2003 AMC 10B Problem 8

Let the terms be $a,$ $ar,$ $ar^2,$ $ar^3,$ $\dots$ with $ar=2$ and $ar^3=6.$ Dividing gives $r^2=3,$ so $r=\pm\sqrt3.$ Then $a=\dfrac{2}{r}=\pm\dfrac{2}{\sqrt3}=\pm\dfrac{2\sqrt3}{3}.$ The choice $-\dfrac{2\sqrt3}{3}$ matches the negative case. Thus, the correct answer is B.