已知等比数列的前n项和sn,若Sn,S(n+2),S(n+1)成等差数列,求公比

设首项a1,公比q
①当q≠1时
Sn=a1(1-q^n)/(1-q)
S(n+2)=a1[1-q^(n+2)]/(1-q)
S(n+1)=a1[1-q^(n+1)]/(1-q)
因为Sn,S(n+2),S(n+1)成等差数列
所以2S(n+2)=Sn+S(n+1)
2a1[1-q^(n+2)]/(1-q)=a1(1-q^n)/(1-q)+a1[1-q^(n+1)]/(1-q)
2[1-q^(n+2)]=(1-q^n)+[1-q^(n+1)]
2-2q^(n+2)=2-q^n-q^(n+1)
2q^(n+2)=q^n+q^(n+1)
2q^2*q^n=q^n+q*q^n
2q^2=1+q
q^2-1/2q=1/2
(q-1/4)^2=9/16
q=1(舍去)或-1/2
②当q=1时
Sn=na1
S(n+2)=(n+2)a1
S(n+1)=(n+1)a1
2S(n+2)=(2n+4)a1
Sn+S(n+1)=(2n+1)a1
2S(n+2)≠Sn+S(n+1)因此不成等差数列
所以综合以上,q=-1/2

-1/2