等差數列{an}的前n項和為Sn，若a1=﹣12，S5=S8，則當Sn取得最小值時，n的值為（　　）     ...

問題詳情：

等差數列{an}的前n項和為Sn，若a1=﹣12，S5=S8，則當Sn取得最小值時，n的值為（　　）                 

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

【回答】

C【分析】由等差數列前n項和公式，列出方程求出公差d=2，由此能求出Sn，再利用*法能求出當Sn取得最小值時，n的值．                                                                                                                

【解答】解：∵等差數列{an}的前n項和為Sn，a1=﹣12，S5=S8，                                

∴，                                                            

解得d=2，                                                                                                          

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

∴當Sn取得最小值時，n=6或n=7．                                                                      

故選：C．                                                                                                          

【點評】本題考查等差數列的前n項和取最小值時，n的值的求法，是基礎題，解題時要認真審題，注意等差數列的*質的合理運用．                                                                                                         

知識點：數列

題型：選擇題