已知数列 $\left\{a_{n}\right\}$ 为等差数列, 且 $a_{2}=1$, 则 $2\left|a_{9}\right|+\left|a_{10}\right|$ 的最小值为 ( )
设数列 $\left\{a_{n}\right\}$ 的公差为 $d, \because a_{2}=1, \therefore 2\left|a_{9}\right|+\left|a_{10}\right|=2|1+d|+|1+2 d|=\left\{\begin{array}{l}-3-4 d, d<-1 \\ 1,-1 \leq d \leq-\frac{1}{2} \\ 3+4 d, d>-\frac{1}{2}\end{array}\right.$,由分段函数的性质可得 $2\left|a_{9}\right|+\left|a_{10}\right|$ 的最小值为 1 , 故选 C.