记正项等比数列 $\left\{a_{n}\right\}$ 的前 $n$ 项和为 $S_{n}$, 若 $4\left(a_{1}+a_{2}\right)=3, S_{6}=\frac{63}{4}$, 则 $a_{7}=(\quad)$
解析: 依题意, 得 $S_{6}-a_{1}-a_{2}=15$, 所以 $\left\{\begin{array}{l}a_{1}(1+q)=\frac{3}{4}, \\ a_{1} q^{2}\left(1+q+q^{2}+q^{3}\right)=15 .\end{array}\right.$ 两式相除可得, $q^{2}\left(1+q^{2}\right)=20$.
所以 $\left\{\begin{array}{l}a_{1}=\frac{1}{4}, \text { 所以 } a_{7}=a_{1} q^{6}=\frac{1}{4} \times 2^{6}=16 . \\ q=2 .\end{array}\right.$