设{an}是公比不等于1的等比数列,Sn是其前n项和,若a1,2a7,3a4成等差数列,求证12S3,S6,S12-S6

q=公比≠1
a7=a4q^3
4a7=a1+3a4=4a4q^3
a4=a1/(4q^3-3)=a1q^3,Q=q^3
1=(4Q-3)Q=4Q^2-3Q
4Q^2-3Q-1=0
(4Q+1)(Q-1)=0
Q=-1/4
q=(-1/4)^(1/3)
令z=1-q
12S3=12a1(1-q^3)/(1-q)=15a1/(1-q)=15a1/z
S6=a1(1-1/16)/(1-q)=15a1/16z
S12-S6=(a1/z)[1-1/16^2-1+1/16]
=(a1/z)*15/16^2
S6/12S3=[(a1/z)*15/16]/15(a1/z)=1/16
(S12-S6)S6=(15/16^2)/（15/16）=1/16=S6/12S3
所以,12S3,S6,S12-S6成等比数列