以知数列{An}是等比数列.公比Q不等于1,Sn是其前n项和,a1,a7,a4成等差数列.求证2S3,S6,S12-S6等比

a1,a7,a4成等差数列
2a7=a1+a4
2a1q^6=a1+a1q^3
2q^6=1+q^3
2q^6-q^3-1=(2q^3+1)(q^3-1)=0
因为公比Q不等于1，
所以，q^3=-1/2,

2S3*（S12-S6)
=2a1(1-q^3)/(1-q)*[a1(1-q^12)/(1-q)-a1(1-q^6)/(1-q)]
=2a1(1+1/2)/(1-q)*[a1(1-1/16)/(1-q)-a1(1-1/4)/(1-q)]
=[a1/(1-q)]^2[3*(15/16-3/4)
=[a1/(1-q)]^2*9/16
=[a1*(3/4)/(1-q)]^2
=[a1*(1-1/4)/(1-q)]^2
=[a1*(1-q^6)/(1-q)]^2
=S6^2
2S3,S6,S12-S6等比