设等比数列an的公比为q,前n项和为sn,若s(n+1),sn,s(n+2)成等差数列,求q的值

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若q=1,则S(n+1)=n+1,Sn=n,S(n+2)=n+2,
此时S(n+1),Sn,S(n+2)不成等差数列
所以q≠1,则Sn=a1*(1-q^n)/(1-q)
a1*[1-q^(n+1)]/(1-q)+a1*[1-q^(n+2)]/(1-q)=2*a1*(1-q^n)/(1-q)
[1-q^(n+1)]+[1-q^(n+2)]=2-2*q^n
q^(n+1)+q^(n+2)=2*q^n
q+q²=2
q²+q-2=0
(q+2)(q-1)=0
∴q=-2

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