设等比数列{an}的公比为q,前n项和为Sn,若Sn+1,Sn,Sn+2成等差数列,则公比q为( )
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设等比数列{an}的公比为q,前n项和为Sn,若Sn+1,Sn,Sn+2成等差数列,则公比q为( )
A. q=-2
B. q=1
C. q=-2或q=1
D. q=2或q=-1
A. q=-2
B. q=1
C. q=-2或q=1
D. q=2或q=-1
设等比数列{an}的公比为q,前n项和为Sn,且Sn+1,Sn,Sn+2成等差数列,则2Sn=Sn+1+Sn+2 .
若q=1,则Sn=na1,式子显然不成立.
若q≠1,则有 2
a1(1−qn)
1−q=
a1(1−qn+1)
1−q+
a1(1−qn+2)
1−q,
故2qn=qn+1+qn+2,即q2+q-2=0,因此q=-2.
故选:A.
若q=1,则Sn=na1,式子显然不成立.
若q≠1,则有 2
a1(1−qn)
1−q=
a1(1−qn+1)
1−q+
a1(1−qn+2)
1−q,
故2qn=qn+1+qn+2,即q2+q-2=0,因此q=-2.
故选:A.
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