2008 AMC 12A Problems/Problem 16

Revision as of 15:51, 17 February 2008 by Xantos C. Guin (talk | contribs) (New page: ==Problem== The numbers <math>\log(a^3b^7)</math>, <math>\log(a^5b^{12})</math>, and <math>\log(a^8b^{15})</math> are the first three terms of an arithmetic sequence, and the <math>12^\te...)
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Problem

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

$\textbf{(A)}\ 40 \qquad \textbf{(B)}\ 56 \qquad \textbf{(C)}\ 76 \qquad \textbf{(D)}\ 112 \qquad \textbf{(E)}\ 143$

Solution

Let $A = \log(a)$ and $B = \log(b)$.

The first three terms of the arithmetic sequence are $3A + 7B$, $5A + 12B$, and $8A + 15B$, and the $12^\text{th}$ term is $nB$.

Thus, $2(5A + 12B) = (3A + 7B) + (8A + 15B) \Rightarrow A = 2B$.

Since the first three terms in the sequence are $13B$, $22B$, and $31B$, the $k$th term is $(9k + 4)B$.

Thus the $12^\text{th}$ term is $(9\cdot12 + 4)B = 112B = nB \Rightarrow n = 112\Rightarrow D$