（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-S

来源：学生作业帮编辑：作业帮分类：数学作业时间：2024/05/23 02:04:16

（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

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.

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 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 已知数列{An}是等差数列,Sn是其前n项的和,求证S6,S12-S6,S18-S12也成等差数列. 已知数列{An}是等差数列,Sn是其前n项的和,求证S6, S12-S6,S18-S12也成等差数列. 辛苦, 多谢在已知数列{an}是首项为a且公比不等于1的等比数列,Sn为其前n项和,a1,2a7,3a4成等差数列,求证：12S3,S 已知数列an是首项为a 且公比q不等于一1的等比数列 sn是其前n项和 a1 2a7 3a4成等差数列已知数列an是首项为a且公比q不等于1的等比数列,Sn是其前n项和,a1,2a7,3a4成等差数列. 已知数列{an}是等差数列,Sn是其前几项的和,求证S6,S12-S6,S18-S12也成等差数列已知数列{An}是首项为a且公比q不等于1得等比数列,Sn是其前n项和,A1,2A7,3A4成等差数列. 已知Sn是等比数列{An}的前N项和,S3,S9,S6成等差数列,则设等比数列{an}的前n项和为sn,若S6,S9,S3成等差数列,问2S3,S6,S12-S6S能否成等比数列?请说明理等比数列证明等比数列首项为a,公比q不为0,Sn为前n项和,a1,2a7,3a4成等差数列.证明12S3,S6,S12-