2003 AMC 12B Problem 6 — Solution | LIVE by Po-Shen Loh

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2003 AMC 12B Problem 6

Attempt Problem 6 of the 2003 AMC 12B below, then check your answer against the professionally curated solution from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2003 AMC 12B solutions, or check the answer key.

All problems are used with official legal permission of the Mathematical Association of America (MAA).

6.

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

$-\sqrt{3}$

$-\dfrac{2\sqrt{3}}{3}$

$-\dfrac{\sqrt{3}}{3}$

$\sqrt{3}$

$3$

Answer: B

Concepts:geometric sequence radical

Difficulty rating: 1290

Solution:

Let the first term be $a$ and the common ratio $r.$ Then $ar = 2$ and $ar^3 = 6,$ so $r^2 = 3$ and $r = \pm\sqrt{3}.$

The first term is $a = \frac{2}{r} = \pm\frac{2}{\sqrt{3}} = \pm\frac{2\sqrt{3}}{3}.$ The choice $-\dfrac{2\sqrt{3}}{3}$ appears among the options.

Thus, the correct answer is B.

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