Question

（课129 7)已知数列{an}是等比数列，Sn是其前n项的和，a1,a7,a4成等差数列，求证2S3,S6,S12-S6成等比数列

Answer 1

a(7)=a(1)(q^6)
a(4)=a(1)(q^3)
则依题意可得
2a(7)=a(1)+a(4)
所以
2a(1)(q^6)=a(1)+a(1)(q^3)
化简得
2(q^6)-(q^3)-1=0
解得
q^3=1或q^3=-1/2。
对于2S(3)、S(6)和S(12)-S(6)
当它们成等比数列时，
应满足S(6)²=2S(3)×[S(12)-S(6)]
显然，
若q^3=1，q=1，
则
2S(3)=2×3a(1)=6a(1)
S(6)=6a(1)
S(12)-S(6)=6a(1)
上述三者相等，其等比数列关系成立；
当q^3=-1/2时，q^6=1/4，q^12=1/16
则
2S(3)=2a(1)[1-(q^3)]/(1-q)=3a(1)/(1-q)
S(6)=a(1)[1-(q^6)]/(1-q)=(3/4)a(1)/(1-q)
S(12)-S(6)=a(1)[1-(q^12)]/(1-q)-a(1)[1-(q^6)]/(1-q)
=(15/16-3/4)a(1)/(1-q)
=(3/16)a(1)/(1-q)
可见，
2S(3)×[S(12)-S(6)]=(9/16)[a(1)²/(1-q)²] =S(6)²
即2S(3)、S(6)和S(12)-S(6)成等比数列。

Answer 2

证明:设数列{an}的公比为q，则a1=a4/q^3, a7=a4q^3,
又因为a1,a7,a4成等差数列,故2a4q^3=a4/q^3+a4,解得q^3=-1/2 或 q^3=1
S6/2S3=[a1(1-q^6)/(1-q)]/2[a1(1-q^3)/(1-q)]=(1-q^6)/2(1-q^3)=(1+q^3)/2,
(S12-S6)/S6=S12/S6-1=(1-q^12)/(1-q^6)-1=q^6
当q^3=-1/2时,S6/2S3=(1+q^3)/2=(1-1/2)/2=1/4,(S12-S6)/S6=q^6=1/4;
当q^3=1时,S6/2S3=(1+q^3)/2=(1+1)/2=1,(S12-S6)/S6=q^6=1
故S6/2S3=(S12-S6)/S6
即证2S3,S6,S12-S6成等比数列