已知等差数列\(\left\{ {{a_n}} \right\}\)的前n项和为\({S_n}\)，若\({S_{12}} = 3{a_{11}}\)，则\({a_5} = \)___，\({S}_{9}=\)___．

等差数列\(\left\{ {{a_n}} \right\}\)中，\({S_{12}} = 12{a_1} + 66d = 3{a_{11}} = 3{a_1} + 30d\)，

所以\(12{a_1} + 66d = 3{a_1} + 30d\)，

即\({a_1} = - 4d\)，

所以\({a_5} = {a_1} + 4d = 0\)，\({S_9} = 9{a_5} = 0\)

故答案为：①0；②0.

进入考试题库