2003 AMC 10B Problem 8

1. In a geometric sequence you multiply by the same number $r$ (the ratio) to get each next term. From the 2nd term to the 4th, you multiply by $r$ twice:
$$2 \times r \times r = 6, \quad\text{so}\quad r^2 = 3.$$
2. Two numbers square to $3$: $r = \sqrt{3}$ and $r = -\sqrt{3}$. Both are allowed, and the word "possible" in the question hints that more than one first term works.
3. The first term times $r$ gives the second term, $2$. So the first term is $2 \div r$:
$$\frac{2}{\sqrt{3}} \quad\text{or}\quad -\frac{2}{\sqrt{3}}.$$
4. The choices have no square root on the bottom, so rewrite: multiply top and bottom by $\sqrt{3}$ to get $\frac{2}{\sqrt{3}} = \frac{2\sqrt{3}}{3}$. The two possible first terms are $\frac{2\sqrt{3}}{3}$ and $-\frac{2\sqrt{3}}{3}$.
5. Only $-\frac{2\sqrt{3}}{3}$ is in the list. Check with $r = -\sqrt{3}$: the sequence is $-\frac{2\sqrt{3}}{3},\ 2,\ -2\sqrt{3},\ 6$. The 2nd term is $2$ and the 4th is $6$, as required.

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

Two terms that are $k$ places apart differ by a factor of $r^k$, so they tell you $r^k$ without knowing the first term. Here only $r^2$ is known, and both signs of $r$ fit. Before comparing with the choices, write your answer in the same form they use.

Call the first term $a$. Then $a \cdot r = 2$, so $a^2 \cdot r^2 = 4$. Since $r^2 = 3$, this gives $a^2 = \frac{4}{3}$. Now square each choice: only (B) gives $\frac{4 \cdot 3}{9} = \frac{4}{3}$.

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