2008 AMC 12A Problem 16

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, $$\begin{aligned} &2\log a + 5\log b \\ &= 3\log a + 3\log b, \end{aligned}$$ 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 $$\begin{aligned} (13 + 11 \cdot 9)\log b &= 112\log b \\ &= \log(b^{112}), \end{aligned}$$ so $n = 112.$ Thus, D is the correct answer.