设等差数列 $\left\{a_{n}\right\}$ 的前 $n$ 项和为 $S_{n}$, 若 $S_{3}=9, S_{6}=36$. 则 $a_{7}+a_{8}+a_{9}$ 等于 ( )
解析 数列 $\left\{a_{n}\right\}$ 为等差数列, 则 $S_{3}, S_{6}-S_{3}, S_{9}-S_{6}$ 为等差数列, 即 $2\left(S_{6}-S_{3}\right)=S_{3}+\left(S_{9}\right.$ $-S_{6}$ ),

$\because S_{3}=9, S_{6}-S_{3}=27$, 则 $S_{9}-S_{6}=45$.

$\therefore a_{7}+a_{8}+a_{9}=S_{9}-S_{6}=45$.