已知等比数列 $\left\{a_{n}\right\}$ 满足 $a_{1}+a_{2}=3, a_{2}+a_{3}=6$, 则 $a_{7}$ 等于 ( )
解析 $\because\left\{a_{n}\right\}$ 为等比数列,

$\therefore \frac{a_{2}+a_{3}}{a_{1}+a_{2}}=q=2$.

又 $a_{1}+a_{2}=3, \therefore a_{1}=1$. 故 $a_{7}=1 \cdot 2^{6}=64$.