设等差数列\(\left\{ {{a_n}} \right\}\)的前\(n\)和为\({S_n},\)且\({S_{m - 1}} = - 4,{S_m} = 0,{S_{m + 1}} = 5,\)则\(m = \)___
由题意得：\({a_m} = {S_m} - {S_{m - 1}} = 4\)，\({a_{m + 1}} = {S_{m + 1}} - {S_m} = 5\)，
则等差数列的公差\(d = {a_{m + 1}} - {a_m} = 1\)，
则\({a_m} = {a_1} + \left( {m - 1} \right)d = {a_1} + \left( {m - 1} \right) = 4\)，\({S_m} = {a_1}m + \frac{{m\left( {m - 1} \right)}}{2} = 0\)，
解得：\(m = 9\)或\(m = 0\)（舍去），\({a_1} = - 4\)，
故答案为：9
