12. In the sequence $\left\{a_{n}\right\}$, $a_{1}=10$, and $a_{n+1}=2 \sqrt{a_{n}}$, then its general term formula $a_{n}=$
12. $a_{n}=4 \cdot\left(\frac{5}{2}\right)^{2^{1-n}}\left(n \in \mathbf{N}^{*}\right)$.

Brief solution: From the given, we have $\lg a_{n+1}=\frac{1}{2} \lg a_{n}+\lg 2$. Let $b_{n}=\lg a_{n}$, then $b_{n+1}-2 \lg 2=\frac{1}{2}\left(b_{n}-2 \lg 2\right)$. Thus, $b_{n}-2 \lg 2=\left(\frac{1}{2}\right)^{n-1}\left(b_{1}-2 \lg 2\right)$. From $b_{1}=1$, we get $b_{n}=\lg 4+\left(\frac{1}{2}\right)^{n-1} \lg \frac{5}{2}$, hence $a_{n}=4 \cdot$ $\left(\frac{5}{2}\right)^{2^{1-n}}\left(n \in \mathbf{N}^{*}\right)$