2003 AMC 10B Problem 8 — Solution | LIVE by Po-Shen Loh

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2003 AMC 10B Problem 8

Below is the professionally curated solution for Problem 8 of the 2003 AMC 10B, from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2003 AMC 10B solutions, or check the answer key.

All of the real AMC 8, AMC 10, AMC 12, and AIME problems in our complete solution collection are used with official legal permission of the Mathematical Association of America (MAA).

Concepts:geometric sequence radical

Difficulty rating: 1140

8.

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 terms be $a, ar, ar^2, ar^3,\dots$ with $ar=2$ and $ar^3=6.$ Dividing gives $r^2=3,$ so $r=\pm\sqrt3.$

Then $a=\dfrac{2}{r}=\pm\dfrac{2}{\sqrt3}=\pm\dfrac{2\sqrt3}{3}.$ The choice $-\dfrac{2\sqrt3}{3}$ matches the negative case.

Thus, the correct answer is B.

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