jmc

The common ratio of the geometric sequence is $\frac{16}{12}=\frac{4}{3}$. Thus, the $n^{\text{th}}$ term of the sequence is given by $a{\left(\frac{4}{3}\right)}^{n-1}$, where $a$ is the first term of the sequence. Since the third term is 12, we plug in $n=3$ to get $a{\left(\frac{4}{3}\right)}^2=12$. Solving, we have $a\left(\frac{16}{9}\right)=12\Rightarrow a=12\left(\frac{9}{16}\right)=\boxed{\frac{27}{4}}$.