AMC12 2008 12A, Problem 16

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 $n$?

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, $2(5A + 12B) = (3A + 7B) + (8A + 15B)  = 2B$. 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 $(9\cdot12 + 4)B = 112B = nB \Rightarrow n = 112\Rightarrow \fbox{112}$.

Answer: 112