12. 设 $\left\{a_{n}\right\}$ 为等差数列, $S_{n}$ 为数列 $\left\{a_{n}\right\}$ 的前 $n$ 项和, 已知 $S_{7}=7, S_{15}=75, T_{n}$ 为数列 $\left\{\frac{S_{n}}{n}\right\}$的前 $n$ 项和, 求 $T_{n}$.
设等差数列 $\left\{a_{n}\right\}$ 的公差为 $d$,

则 $S_{n}=n a_{1}+\frac{1}{2} n(n-1) d$,

$\because S_{7}=7, S_{15}=75, \quad \therefore\left\{\begin{array}{l}7 a_{1}+21 d=7 \\ 15 a_{1}+105 d=75\end{array}\right.$,

即 $\left\{\begin{array}{l}a_{1}+3 d=1 \\ a_{1}+7 d=5\end{array}\right.$, 解得 $\left\{\begin{array}{l}a_{1}=-2 \\ d=1\end{array}\right.$,

$\therefore \frac{S_{n}}{n}=a_{1}+\frac{1}{2}(n-1) d=-2+\frac{1}{2}(n-1)$,

$\because \frac{S_{n+1}}{n+1}-\frac{S_{n}}{n}=\frac{1}{2}$,

$\therefore$ 数列 $\left\{\frac{S_{n}}{n}\right\}$ 是等差数列, 其首项为 -2 , 公差为 $\frac{1}{2}$,

$\therefore T_{n}=n \times(-2)+\frac{n(n-1)}{2} \times \frac{1}{2}=\frac{1}{4} n^{2}-\frac{9}{4} n$.