数列 $\left\{a_{n}\right\}$ 满足 $a_{n+1}-a_{n}=2\left(n \in N^{*}\right)$, 且 $a_{2}+a_{40}=680$, 则 $a_{22}=(\quad)$
数列 $\left\{a_{n}\right\}$ 满足 $a_{n+1}-a_{n}=2\left(n \in N^{*}\right)$, 且 $a_{2}+a_{40}=680$,
$\therefore$ 数列 $\left\{\mathrm{a}_{n}\right\}$ 是公差为 $\mathrm{d}=2$ 的等差数列, $\therefore \mathrm{a}_{3}+\mathrm{a}_{40}=\mathrm{a}_{1}+2+\mathrm{a}_{1}+39 \times 2=680$,

解得 $\mathrm{a}_{1}=300, \therefore \mathrm{a}_{22}=299+21 \times 2=342$. 故选 $: C$.