已知数列｛a}是等比数列,Sn是其前n项和,a1,a7,a4成等差数列,求证2S3,S6,S12－S6成等比数列

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a7=a1q^6
a4=a1q^3
因为a1,a7,a4成等差数列
所以a7=(a1+a4)/2
a1q^6=(a1+a1q^3)/2
2q^6-q^3-1=0
(2q^3+1)(q^3-1)=0
q^3=1,或q^3=-1/2
因为
2S3=2a1(1-q^3)/(1-q)
S6=a1(1-q^6)/(1-q)
S12-S6=a1(1-q^12)/(1-q)-a1(1-q^6)/(1-q)
S6/2S3=(1+q^3)/2
(S12-S6)/S6=(1+q^6)-1
当q^3=1时,S6/2S3=(1+q^3)/2=1,(S12-S6)/S6=(1+q^6)-1
=1
当q^3=-1/2时,S6/2S3=(1+q^3)/2=1/4,(S12-S6)/S6=(1+q^6)-1
=1/4
所以2S3,S6,S12－S6成等比数列

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