已知{an}为等差数列，a1+a3+a5=105，a2+a4+a6=99，则a20等于（ ）
A. -1
B. 1
C. 3
D. 7
人气：338 ℃ 时间：2019-12-13 01:36:12
解答
由已知得a1+a3+a5=3a3=105，
a2+a4+a6=3a4=99，
∴a3=35，a4=33，∴d=a4-a3=-2．
∴a20=a3+17d=35+（-2）×17=1．
故选B
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