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q的值为

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2Sn=S(n+2)+S(n+1)(q不为1)
2a1(1-q^n)/(1-q)=a1[1-q^(n+2)]/(1-q)+a1[1-q^(n+1)]/(1-q)
2-2q^n=2-[q^(n+1)+q^(n+2)]
那么q+q^2=0
可得q=-1或q=0(舍去)
当q=1,Sn=na1
Sn+1=(n+1)a1
S(n+2)=(n+2)a1
显然a1=0,.
那就不可以了.
所以q=-1

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