2013 AMC 12A Problem 14

Below is the professionally curated solution for Problem 14 of the 2013 AMC 12A, from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2013 AMC 12A solutions, or check the answer key.

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Concepts:logarithm geometric sequence arithmetic sequence

Difficulty rating: 1800

14.

The sequence $\log_{12} 162, \ \log_{12} x, \ \log_{12} y, \ \log_{12} z, \ \log_{12} 1250$ is an arithmetic progression. What is $x?$

$125\sqrt{3}$

$270$

$162\sqrt{5}$

$434$

$225\sqrt{6}$

Solution:

Because the logarithms are in arithmetic progression, $162, x, y, z, 1250$ is a geometric sequence. Its common ratio $r$ satisfies $162 r^4 = 1250,$ so $r^4 = \tfrac{625}{81}$ and $r = \tfrac53.$

Therefore $x = 162\cdot\tfrac53 = 270.$

Thus, the correct answer is B.

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