2003 AMC 10B Problem 8

Let the common ratio be $r$. Going from the second term to the fourth multiplies by $r$ twice, so
$$r^2 = \frac{6}{2} = 3, \qquad r = \pm\sqrt{3}.$$

The first term is the second term divided by $r$:
$$\frac{2}{\pm\sqrt{3}} = \pm\frac{2}{\sqrt{3}} = \pm\frac{2\sqrt{3}}{3}.$$

Of these two possibilities, only $-\frac{2\sqrt{3}}{3}$ appears among the choices. Check: with $r = -\sqrt{3}$ the sequence begins $-\frac{2\sqrt3}{3},\ 2,\ -2\sqrt{3},\ 6$, as required.

The answer is $\boxed{\textbf{(B)}\ -\frac{2\sqrt{3}}{3}}$.

In a geometric sequence, terms $k$ places apart are related by $r^k$, so two given terms determine $r^k$ without knowing the first term. Because only $r^2$ is known, both signs of $r$ are legitimate, and the question's word "possible" signals that more than one first term works. Always rationalize before comparing with answer choices written in the form $\frac{a\sqrt{b}}{c}$.

Test choices: the first term $a$ must satisfy $a r = 2$ and $a r^3 = 6$, so $a^2 r^2 = 4$ and $r^2 = 3$ give $a^2 = \frac{4}{3}$. Only choice (B) squares to $\frac{4}{3}$.

• Forgetting that $r$ can be negative and rejecting the negative choices, or failing to rationalize $\frac{2}{\sqrt3}$ and not recognizing it among the choices.
• Taking $r = \frac{6}{2} = 3$ as if the terms were consecutive, which gives the first term $\frac{2}{3}$ and matches nothing.