2008 AMC 12A Problem 16

Below is the professionally curated solution for Problem 16 of the 2008 AMC 12A, from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2008 AMC 12A solutions, or check the answer key.

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Concepts:arithmetic sequence logarithm

Difficulty rating: 1800

16.

The numbers $\log(a^3 b^7),$ $\log(a^5 b^{12}),$ and $\log(a^8 b^{15})$ are the first three terms of an arithmetic sequence, and the $12$ th term of the sequence is $\log(b^n).$ What is $n?$

$40$

$56$

$76$

$112$

$143$

Solution:

The three terms are $3\log a + 7\log b,$ $5\log a + 12\log b,$ and $8\log a + 15\log b.$ Setting the two consecutive differences equal, $2\log a + 5\log b = 3\log a + 3\log b,$ so $\log a = 2\log b.$

The first term is then $(3 \cdot 2 + 7)\log b = 13\log b,$ and the common difference is $(2 \cdot 2 + 5)\log b = 9\log b.$

The $12$ th term is $(13 + 11 \cdot 9)\log b = 112\log b = \log(b^{112}),$ so $n = 112.$

Thus, D is the correct answer.

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