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

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2003 AMC 12B Problem 6

Below is the professionally curated solution for Problem 6 of the 2003 AMC 12B, 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.

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Concepts:geometric sequence radical

Difficulty rating: 1290

6.

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

$-\sqrt{3}$

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

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

$\sqrt{3}$

$3$

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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