7. 设 $S_{n}$ 为等差数列 $\left\{a_{n}\right\}$ 的前 $n$ 项和, 若 $S_{3}=3, S_{6}=24$, 则 $a_{9}=$
设等差数列的公差为 $d$, 则

$S_{3}=3 a_{1}+\frac{3 \times 2}{2} d=3 a_{1}+3 d=3$,

即 $a_{1}+d=1$,

$S_{6}=6 a_{1}+\frac{6 \times 5}{2} d=6 a_{1}+15 d=24$,

即 $2 a_{1}+5 d=8$.

由 $\left\{\begin{array}{l}a_{1}+d=1, \\ 2 a_{1}+5 d=8,\end{array}\right.$ 解得 $\left\{\begin{array}{l}a_{1}=-1, \\ d=2 .\end{array}\right.$

故 $a_{9}=a_{1}+8 d=-1+8 \times 2=15$.