14. 设等差数列 $\left\{a_{n}\right\}$ 的前 $n$ 项和为 $S_{n}$, 已知 $a_{3}=12$, 且 $S_{12}>0, S_{13}<0$.

(1)求公差 $d$ 的范围;

(2)问前几项的和最大, 并说明理由.
(1)根据题意, 有: $\left\{\begin{array}{l}12 a_{1}+\frac{12 \times 11}{2} d>0, \\ 13 a_{1}+\frac{13 \times 12}{2} d<0, \\ a_{1}+2 d=12,\end{array} \quad\right.$ 整理得: $\left\{\begin{array}{l}2 a_{1}+11 d>0, \\ a_{1}+6 d<0, \\ a_{1}+2 d=12 .\end{array}\right.$

解之得: $-\frac{24}{7}<d<-3$.

(2) $\because d<0$,

而 $S_{13}=\frac{13\left(a_{1}+a_{13}\right)}{2}=13 a_{7}<0, \quad \therefore a_{7}<0$.

又 $S_{12}=\frac{12\left(a_{1}+a_{12}\right)}{2}=6\left(a_{1}+a_{12}\right)=6\left(a_{6}+a_{7}\right)>0$,

$\therefore a_{6}>0$.

$\therefore$ 数列 $\left\{a_{n}\right\}$ 的前 6 项和 $S_{6}$ 最大.