# Difference between revisions of "2003 AMC 10B Problems/Problem 8"

The following problem is from both the 2003 AMC 12B #6 and 2003 AMC 10B #8, so both problems redirect to this page.

## Problem

The second and fourth terms of a geometric sequence are $2$ and $6$. Which of the following is a possible first term?

$\textbf{(A) } -\sqrt{3} \qquad\textbf{(B) } -\frac{2\sqrt{3}}{3} \qquad\textbf{(C) } -\frac{\sqrt{3}}{3} \qquad\textbf{(D) } \sqrt{3} \qquad\textbf{(E) } 3$

## Solution

Let the first term be $a$ and the common ratio be $r$. Therefore,

$$ar=2\ \ (1) \qquad \text{and} \qquad ar^3=6\ \ (2)$$

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 $\boxed{\textbf{(B)}\ -\frac{2\sqrt{3}}{3}}$ is a possible first term.