已知数列 $\left\{ a_{n}\right\}$ 是各项及公差都不为$0$的等差数列,若 $S_{n}$ 为数列 $\left\{ a_{n}\right\}$ 的前$n$项和,则" $a_{1},a_{2},a_{5}$ 成等比数列"是" $\left\{ \dfrac {S_{n}}{n^{2}}\right\}$ 为常数列"的$ $ $(\qquad)$ $ $

充分不必要条件

必要不充分条件

充要条件

既不充分也不必要条件

设等差数列 $\left\{ a_{n}\right\} $ 的公差为$d$,则 $a_{2}=a_{1}+d,a_{5}=a_{1}+4d$.若 $a_{1},a_{2},a_{5}$ 成等比数列,则 $a_{2}^{2}=a_{1}a_{5}$, 即 $\left(a_{1}+d\right)^{2}=a_{1}\left(a_{1}+4d\right)$, 整理得 $d^{2}=2a_{1}d$, 因为 $d\neq 0$, 所以 $d=2a_{1}$, 所以 $S_{n}=na_{1}+\dfrac {n\left(n-1\right)}{2}d=na_{1}+n\left(n-1\right)a_{1}=n^{2}a_{1}$, 所以 $\dfrac {S_{n}}{n^{2}}=a_{1}$, 所以数列 $\left\{ \dfrac {S_{n}}{n^{2}}\right\} $ 为常数列.若数列 $\left\{ \dfrac {S_{n}}{n^{2}}\right\} $ 为常数列,则 $\dfrac {S_{n}}{n^{2}}=\dfrac {na_{1}+\dfrac {n\left(n-1\right)}{2}d}{n^{2}}=\dfrac {d}{2}n+\left(a_{1}-\dfrac {d}{2}\right)$ 为常数,所以 $a_{1}=\dfrac {d}{2}$, 所以 $a_{2}=a_{1}+d=2a_{1},a_{5}=a_{1}+4d=5a_{1}$, 所以 $a_{2}^{2}=4a_{1}^{2}=a_{1}a_{5}$, 又 $a_{1}\neq 0$, 所以 $a_{1},a_{2},a_{5}$ 成等比数列.综上," $a_{1},a_{2},a_{5}$ 成等比数列"是" $\left\{ \dfrac {S_{n}}{n^{2}}\right\} $ 为常数列"的充要条件.故选 $\rm C $.