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设等差数列{an}的前n项和为Sn,已知(a5-1)3+2009(a5-1)=1,(a2005-1)3+2009(a2005-1)=-1,则下列结论中正确的是()A.S2009=2009,a2005<a5B.S2009=2009,a2005>a5C.S2009=-2009,a2005≤a5D.S2009=-2009
题目详情
设等差数列{an}的前n项和为Sn,已知(a5-1)3+2009(a5-1)=1,(a2005-1)3+2009(a2005-1)=-1,则下列结论中正确的是( )
A. S2009=2009,a2005<a5
B. S2009=2009,a2005>a5
C. S2009=-2009,a2005≤a5
D. S2009=-2009,a2005≥a5
A. S2009=2009,a2005<a5
B. S2009=2009,a2005>a5
C. S2009=-2009,a2005≤a5
D. S2009=-2009,a2005≥a5
▼优质解答
答案和解析
∵等差数列{an}的前n项和为Sn,(a5-1)3+2009(a5-1)=1,
∴(a5-1)[(a5−1)2+2009]=1,
∵(a5−1)2+2009>0,
∴a5-1>0,即:a5>1.
∵(a2005-1)3+2009(a2005-1)=-1,
∴(a2005-1)[(a2005−1)2+2009]=-1,
∵(a2005−1)2+2009>0,
∴a2005-1<0,即:a2005<1.
∴a2005<a5.
据上可知应选A.
故选:A.
∴(a5-1)[(a5−1)2+2009]=1,
∵(a5−1)2+2009>0,
∴a5-1>0,即:a5>1.
∵(a2005-1)3+2009(a2005-1)=-1,
∴(a2005-1)[(a2005−1)2+2009]=-1,
∵(a2005−1)2+2009>0,
∴a2005-1<0,即:a2005<1.
∴a2005<a5.
据上可知应选A.
故选:A.
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