1. The terms of the geometric sequence $\left\{a_{n}\right\}$ are all positive, and $a_{1} a_{3}+a_{2} a_{6}+2 a_{3}^{2}=36$, then the value of $a_{2}+a_{4}$ is . $\qquad$
Answer: 6.
Solution: Since $36=a_{1} a_{3}+a_{2} a_{6}+2 a_{3}^{2}=a_{2}^{2}+a_{4}^{2}+2 a_{2} a_{4}=\left(a_{2}+a_{4}\right)^{2}$, and $a_{2}+a_{4}>0$, thus $a_{2}+a_{4}=6$.