已知等差数列 $\left\{a_{n}\right\}$ 的前 $n$ 项和为 $S_{n}$, 若 $S_{8}=a_{8}=8$, 则公差 $d$ 等于 ( )
解析: $\because S_{8}=a_{8}=8, \therefore a_{1}+a_{2}+\ldots+a_{8}=a_{8}, \therefore S_{7}=0, \because S_{7}=7 a_{4}, \therefore a_{4}=0$

$\because a_{4}=0, a_{8}=8, \therefore d=\frac{a_{8}-a_{4}}{8-4}=2$.