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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=-2B.q=1C.q=-2或q=1D.q=2或q=-1
题目详情
nnn+1nn+2
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答案和解析
设等比数列{ann}的公比为q,前n项和为Snn,且Sn+1n+1,Snn,Sn+2n+2成等差数列,则2Snn=Sn+1n+1+Sn+2 n+2 .
若q=1,则Snn=na11,式子显然不成立.
若q≠1,则有 2
=
+
,
故2qn=qn+1+qn+2,即q2+q-2=0,因此q=-2.
故选:A. 2
a1(1−qn) a1(1−qn) a1(1−qn)1(1−qn)n)1−q 1−q 1−q=
a1(1−qn+1) a1(1−qn+1) a1(1−qn+1)1(1−qn+1)n+1)1−q 1−q 1−q+
a1(1−qn+2) a1(1−qn+2) a1(1−qn+2)1(1−qn+2)n+2)1−q 1−q 1−q,
故2qnn=qn+1n+1+qn+2n+2,即q22+q-2=0,因此q=-2.
故选:A.
若q=1,则Snn=na11,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. 2
a1(1−qn) |
1−q |
a1(1−qn+1) |
1−q |
a1(1−qn+2) |
1−q |
故2qnn=qn+1n+1+qn+2n+2,即q22+q-2=0,因此q=-2.
故选:A.
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