3. Let $\left\{a_{n}\right\}$ be a geometric sequence with real terms, the sum of the first $k$ terms is $S_{k}$, and the common ratio $q$ satisfies $|q| \neq 1$. If $S_{6 n}=$ $2 S_{4 n}+11 S_{2 n}$, then $\frac{S_{10 n}}{S_{8 n}}=$ $\qquad$
3. $\frac{341}{85}$.

Let $S_{k}=A\left(1-B^{k}\right)(A \neq 0,|B| \neq 1)$.
Substitute $S_{6 n}=2 S_{4 n}+11 S_{2 n}$, we get
$1-B^{6 n}=2\left(1-B^{4 n}\right)+11\left(1-B^{2 n}\right)$.
Let $B^{2 n}=x>0$. Then
$1-x^{3}=2\left(1-x^{2}\right)+11(1-x)(x \neq 1)$
$\Rightarrow 1+x+x^{2}=2(1+x)+11$
$\Rightarrow x=4$.
Thus, $\frac{S_{10 n}}{S_{8 n}}=\frac{1-B^{10 n}}{1-B^{8 n}}=\frac{1-x^{5}}{1-x^{4}}=\frac{341}{85}$.