已知等比数列{an}的前n项和为Sn，若Sn，Sn+2，Sn+1成等差数列，则数列{an}的公比为

\u5728\u516c\u6bd4\u4e3aq\u7684\u7b49\u6bd4\u6570\u5217{an}\u4e2d,\u524dn\u9879\u548c\u4e3aSn,\u82e5Sm+2,Sm+1\u6210\u7b49\u5dee\u6570\u5217,\u5219Am,Am+2,Am+1\u6210

\u8bb2\u89e3 \uff081\uff09\u9006\u547d\u9898\uff1a\u5728\u7b49\u6bd4\u6570\u5217{an}\u4e2d\uff0c\u524dn\u9879\u548c\u4e3aSn\uff0c\u82e5am, am+2, am+1\u6210\u7b49\u5dee\u6570\u5217\uff0c\u5219 Sm\uff0cSm+2\uff0cSm+1\u6210\u7b49\u5dee\u6570\u5217.
\uff082\uff09\u8bbe{an}\u7684\u9996\u9879\u4e3aa1\uff0c\u516c\u6bd4\u4e3aq
\u7531\u5df2\u77e5\u5f972am+2= am + am+1 \u22342a1qm+1=a1 +a1qm
\u2235a1\u22600 q\u22600 ,
\u22342q2\uff0dq\uff0d1=0 ,
\u2234q=1\u6216q=\uff0d .
\u5f53q=1\u65f6\uff0c
\u2235Sm=ma1\uff0c Sm+2= (m+2)a1\uff0cSm+1= (m+1)a1\uff0c
\u2234Sm+Sm+1\u22602 Sm+2,
\u2234Sm\uff0cSm+2\uff0cSm+1\u4e0d\u6210\u7b49\u5dee\u6570\u5217.
\u5f53q=\uff0d \u65f6,
2 Sm+2= ,
\u2234Sm+Sm+1=2 Sm+2 ,
\u2234Sm\uff0cSm+2\uff0cSm+1\u6210\u7b49\u5dee\u6570\u5217\uff0e
\u7efc\u4e0a\u5f97\uff1a\u5f53\u516c\u6bd4q=1\u65f6\uff0c\u9006\u547d\u9898\u4e3a\u5047\uff1b
\u5f53\u516c\u6bd4q\u22601\u65f6\uff0c\u9006\u547d\u9898\u4e3a\u771f\uff0e

2S(n+2)=Sn+S(n+1)
\u52192S3=S1+S2
2(a1+a2+a3)=a1+(a1+a2)
\u6240\u4ee5a2=-2a3
q=a3/a2=-1/2
\u6240\u4ee5a3=a1*q²=1/4

简单分析一下，答案如图所示

依题意
2S_n+2=S_n+1+S_n  即2（S_n+1+a_n+2）=S_n+1+（S_n+1-a_n+1）
∴

an+2

an+1

=-

1

2

∴数列{a_n}的公比为-

1

2

故答案为-

1

2

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