2003 AMC 12B Problems/Problem 6

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

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

Solution

Call the first term $a_1$ and the common ratio $r$ . Then the nth term is given by $a_n=a_1r^{n-1}$ . Therefore, $a_2=a_1r$ and $a_4=a_1r^3$ . Substituting 2 and 6 for $a_2$ and $a_4$ respectively gives $2=a_1r$ and $6=a_1r^3$ . Dividing the first equation by the second gives $r^2=3$ , so $r=\pm\sqrt{3}$ Substituting this into the first equation gives $2=a_1\pm\sqrt{3}$ . Dividing by $\pm\sqrt{3}$ and rationalizing the denominator gives $a_1=\pm\frac{2\sqrt{3}}{3}$ . The negative value corresponds to answer choice $\boxed{\text{(B) }\frac{-2\sqrt{3}}{3}}$ The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.

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2003 AMC 12B Problems/Problem 6

Solution