已知等比数列 $\left\{a_{n}\right\}$ 中, 各项都是正数, 且 $a_{1}, \frac{1}{2} a_{3}, 2 a_{2}$ 成等差数列, 则 $\frac{a_{9}+a_{10}}{a_{7}+a_{8}}$ 等于 $(\quad)$
解析 设等比数列 $\left\{a_{n}\right\}$ 的公比为 $q$,

$\because a_{1}, \frac{1}{2} a_{3}, 2 a_{2}$ 成等差数列,

$\therefore a_{3}=a_{1}+2 a_{2}$,

$\therefore a_{1} q^{2}=a_{1}+2 a_{1} q$,

$\therefore q^{2}-2 q-1=0$,

$\therefore q=1 \pm \sqrt{2}$.

$\because a_{n}>0, \quad \therefore q>0, \quad q=1+\sqrt{2}$.

$\therefore \frac{a_{9}+a_{10}}{a_{7}+a_{8}}=q^{2}=(1+\sqrt{2})^{2}=3+2 \sqrt{2}$.