{an}是等比数列,sn为前n项和,若S4,S10,S7成等差数列,证明a1,a7,a4也成等差数列.
证明：设an=a1*q^(n-1),a1≠0,q≠0
若公比q=1,则由S4,S10,S7成等差数列得2*10a1=4a1+7a1,得a1=0,{an}不是等比数列.那么必有q≠1.于是根据S4,S10,S7成等差数列得
2S10=S4+S7
2a1*(1-q^10)/(1-q)=a1*(1-q^4)/(1-q)+a1*(1-q^7)/(1-q)
两边都乘以1-q,得
2a1-2a1*q^10=a1-a1*q^4+a1-a1*q^7
2a1*q^10=a1*q^4+a1*q^7
因q≠0,两边都除以q^4得
2a1*q^6=a1+a1*q^3
也即2a7=a1+a4
故a1,a7,a4也成等差数列.