等比数列AN公比为Q前N项和为SN,若Sn+1,Sn,Sn+2成等差数列,求Q

等比数列AN公比为Q前N项和为SN,若Sn+1,Sn,Sn+2成等差数列,求Q
等比数列AN公比为Q前N项和为SN,若Sn+1,Sn,Sn+2成等差数列,求Q

等比数列AN公比为Q前N项和为SN,若Sn+1,Sn,Sn+2成等差数列,求Q
公比为1时显然不成立.故有：
Sn=a1(1-q^n)/(1-q)
Sn+1=a1[1-q^(n+1)]/(1-q)
Sn+2=a1[1-q^(n+2)]/(1-q)
Sn+1,Sn,Sn+2成等差数列,故有
Sn+1 + Sn+2=2Sn
即2a1(1-q^n)/(1-q)=a1[1-q^(n+1)]/(1-q)+a1[1-q^(n+2)]/(1-q)
化简得 2-2q^n=2-q^(n+1)-q^(n+2)
q^2+q-2=0
解出q=-2

(1+根号5)/2 或(1-根号5)/2