smc

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 $(9\cdot 12+4)B=112B=nB\Rightarrow n=112\Rightarrow \boxed{112}$.