等差数列{an}的前n项和为Sn，若a1=﹣12，S5=S8，则当Sn取得最小值时，n的值为（　　）     ...

问题详情：

等差数列{an}的前n项和为Sn，若a1=﹣12，S5=S8，则当Sn取得最小值时，n的值为（　　）                 

A．6                                  B．7                                  C．6或7                           D．8

【回答】

C【分析】由等差数列前n项和公式，列出方程求出公差d=2，由此能求出Sn，再利用*法能求出当Sn取得最小值时，n的值．                                                                                                                

【解答】解：∵等差数列{an}的前n项和为Sn，a1=﹣12，S5=S8，                                

∴，                                                            

解得d=2，                                                                                                          

∴Sn=﹣12n+=n2﹣13n=（n﹣）2﹣，                                          

∴当Sn取得最小值时，n=6或n=7．                                                                      

故选：C．                                                                                                          

【点评】本题考查等差数列的前n项和取最小值时，n的值的求法，是基础题，解题时要认真审题，注意等差数列的*质的合理运用．                                                                                                         

知识点：数列

题型：选择题