2. Given positive numbers $a_{1}, a_{2}, \cdots, a_{7}$ form a geometric sequence. If the sum of the first 5 terms is $7 \sqrt{2}+6$, and the sum of the last 5 terms is $14 \sqrt{2}+12$, then $a_{6}$ equals ( )
(A) 4
(B) $4 \sqrt{2}$
(C) 8
(D) $8 \sqrt{2}$
2.C.

Because $a_{1}+a_{2}+a_{3}+a_{4}+a_{5}$
$$
\begin{array}{l}
=a_{1}\left(1+q+q^{2}+q^{3}+q^{4}\right)=7 \sqrt{2}+6, \\
a_{3}+a_{4}+a_{5}+a_{6}+a_{7} \\
=a_{1} q^{2}\left(1+q+q^{2}+q^{3}+q^{4}\right)=14 \sqrt{2}+12,
\end{array}
$$

Dividing the second equation by the first, we get $q^{2}=2$.
Since $a_{1}, a_{2}, \cdots, a_{1}$ are positive, then $q=\sqrt{2}$. Substituting into the above equation, we get $a_{1}=\sqrt{2}$. Therefore, $a_{6}=a_{1} q^{5}=8$.