2014 AMC 12A Problem 7

Below is the professionally curated solution for Problem 7 of the 2014 AMC 12A, from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2014 AMC 12A solutions, or check the answer key.

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Concepts:geometric sequenceexponent

Difficulty rating: 1340

7.

The first three terms of a geometric progression are 3,\sqrt{3}, 33,\sqrt[3]{3}, and 36.\sqrt[6]{3}. What is the fourth term?

11

37\sqrt[7]{3}

38\sqrt[8]{3}

39\sqrt[9]{3}

310\sqrt[10]{3}

Solution:

Writing the terms as powers of 3,3, they are 31/2,3^{1/2}, 31/3,3^{1/3}, 31/6.3^{1/6}. The common ratio is 31/331/2=31/6.\dfrac{3^{1/3}}{3^{1/2}}=3^{-1/6}.

The fourth term is 31/631/6=30=1.3^{1/6}\cdot3^{-1/6}=3^{0}=1.

Thus, the correct answer is A.

Problem 7 in Other Years