已知等比数列 $\left\{a_{n}\right\}$ 的各项都是正数, 且 $3 a_{1}, \frac{1}{2} a_{3}, 2 a_{2}$ 成等差数列, 则 $\frac{a_{8}+a_{9}}{a_{6}+a_{7}}=(\quad)$
: $\therefore 3 a_{10}, \frac{1}{2} a_{3}, 2 a_{2}$ 成等差数列, $\therefore a_{3}=3 a_{1}+2 a_{2}$,

$\therefore q^{2}-2 q-3=0, \therefore q=3$ 或 $q=-1$ (.舍去). $\therefore \frac{a_{8}+a_{9}}{a_{6}+a_{7}^{n}}=\frac{a_{1} q^{7}+a_{1} q^{8}}{a_{1} q^{5}+a_{1} q^{6}}=\frac{q^{2}+q^{3}}{1+q}=q^{2}=3^{2}=9$​​.