∵等差数列{an}的前三项为x-1,x+1,2x+3,
∴(x+1)-(x-1)=(2x+3)-(x+1),
解得x=0.
∴a1=-1,d=2,
an=-1+(n-1)×2=2n-3.
故选B.
若等差数列{an}的前三项为x-1,x+1,2x+3,则这数列的通项公式为( ) A.an=2n-5 B.an=2n-3 C.an=2n-1 D.an=2n+1
若等差数列{an}的前三项为x-1,x+1,2x+3,则这数列的通项公式为( )
A. an=2n-5
B. an=2n-3
C. an=2n-1
D. an=2n+1
A. an=2n-5
B. an=2n-3
C. an=2n-1
D. an=2n+1
数学人气:726 ℃时间:2020-01-27 02:10:47
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