等差数列 $\left\{a_{n}\right\}$ 的公差 $d<0$, 且 $a_{2} \cdot a_{4}=12, a_{2}+a_{4}=8$, 则数列 $\left\{a_{n}\right\}$ 的通项公式是 $(\quad)$
由 $\left\{\begin{array}{l}a_{2} \cdot a_{4}=12, \\ a_{2}+a_{4}=8, \\ d<0,\end{array} \Rightarrow\left\{\begin{array}{l}a_{2}=6, \\ a_{4}=2,\end{array} \Rightarrow\left\{\begin{array}{l}a_{1}=8, \\ d=-2,\end{array}\right.\right.\right.$

所以 $a_{n}=a_{1}+(n-1) d$, 即 $a_{n}=8+(n-1) \times(-2)$,得 $a_{n}=-2 n+10$.