等差数列\(\left\{ {{a_n}} \right\}\)的前\(n\)项和为\({S_n}\)，已知\({S_{10}} = 0\)，\({S_{15}} = 25\)，则\(n + {S_n}\)的最小值为___．

由\({S_n} = n{a_1} + \frac{{n\left( {n - 1} \right)d}}{2}\)，\({S_{10}} = 0\)，\({S_{15}} = 25\)得\(\left\{ {\begin{array}{*{20}{l}}
{10{a_1} + 45d = 0} \\\
{15{a_1} + 105d = 25}
\end{array}} \right.\)，

解得：\(\left\{ {\begin{array}{*{20}{l}}
{{a_1} = - 3} \\\
{d = \frac{2}{3}}
\end{array}} \right.\)，

则\({S_n} = - 3n + \frac{{n\left( {n - 1} \right)}}{2} \cdot \frac{2}{3}{\rm{ = }}\frac{1}{3}\left( {{n^2} - 10n} \right)\)．故\(n + {S_n} = \frac{1}{3}\left( {{n^2} - 7n} \right) = \frac{1}{3}\left[ {{{\left( {n - \frac{7}{2}} \right)}^2} - \frac{{49}}{4}} \right]\)．

由于\(n \in {{\rm{N}}^ * }\)，故当\(n = 3\)或4时，\({\left( {n + {S_n}} \right)_{\min }} = - 4\).

故答案为：\( - 4\)

进入考试题库