考点规范练24等比数列及其前n项和一、基础巩固1.已知等比数列{an}满足a1=14,a3a5=4(a4-1),则a2=()A.2B.1C.12D.182.已知数列{an}的前n项和Sn=Aqn+B(q≠0),则“A=-B”是“数列{an}是等比数列”的()A.充分不必要条件B.必要不充分条件C.充要条件D.既不充分也不必要条件3.设首项为1,公比为23的等比数列{an}的前n项和为Sn,则()A.Sn=2an-1B.Sn=3an-2C.Sn=4-3anD.Sn=3-2an4.已知数列{an}是等比数列,Sn为其前n项和,若a1+a2+a3=4,a4+a5+a6=8,则S12=()A.40B.60C.32D.505.等差数列{an}的首项为1,公差不为0.若a2,a3,a6成等比数列,则{an}前6项的和为()A.-24B.-3C.3D.86.设数列{an}是首项为a1,公差为-1的等差数列,Sn为其前n项和.若S1,S2,S4成等比数列,则a1的值为.7.若等差数列{an}和等比数列{bn}满足a1=b1=-1,a4=b4=8,则a2b2=.8.在等比数列{an}中,an>0,a5-a1=15,a4-a2=6,则a3=.9.在公差不为零的等差数列{an}中,a1=1,a2,a4,a8成等比数列.(1)求数列{an}的通项公式;(2)设bn=2an,Tn=b1+b2+…+bn,求Tn.10.已知等差数列{an}的前n项和为Sn,且S4=4(a3+1),3a3=5a4,数列{bn}是等比数列,且b1b2=b3,2b1=a5.(1)求数列{an},{bn}的通项公式;(2)求数列{|an|}的前n项和Tn.11.在数列{an}中,Sn为数列{an}的前n项和,且Sn=1+kan(k≠0,且k≠1).(1)求an;(2)当k=-1时,求a12+a22+…+an2的值.二、能力提升12.设等差数列{an}的公差d≠0,且a2=-d,若ak是a6与ak+6的等比中项,则k=()A.5B.6C.9D.1113.已知数列{an}的前n项和为Sn,且对任意正整数n都有an=34Sn+2成立.若bn=log2an,则b1008=()A.2017B.2016C.2015D.201414.设等比数列{an}满足a1+a3=10,a2+a4=5,则a1a2…an的最大值为.15.在数列{an}中,a1=2,an+1=n+12n·an(n∈N*).(1)证明:数列{ann}是等比数列,并求数列{an}的通项公式;(2)设bn=an4n-an,若数列{bn}的前n项和是Tn,求证:Tn<2.三、高考预测16.已知数列{an}满足a1=5,a2=5,an+1=an+6an-1(n≥2).(1)求证:{an+1+2an}是等比数列;(2)求数列{an}的通项公式.考点规范练24等比数列及其前n项和1.C解析 a3a5=4(a4-1),∴a42=4(a4-1),解得a4=2.又a4=a1q3,且a1=14,∴q=2.∴a2=a1q=12.2.B解析若A=B=0,则Sn=0,故数列{an}不是等比数列;若数列{an}是等比数列,则a1=Aq+B,a2=Aq2-Aq,a3=Aq3-Aq2,由a3a2=a2a1,得A=-B.3.D解析Sn=a1(1-qn)1-q=a1-anq1-q=1-23an1-23=3-2an,故选D.4.B解析由等比数列的性质可知,数列S3,S6-S3,S9-S6,S12-S9是等比数列,即数列4,8,S9-S6,S12-S9是等比数列,因此S12=4×23+S9=32+4×22+S6=32+16+4+8=60,选B.5.A解析设等差数列的公差为d,则d≠0,a32=a2·a6,即(1+2d)2=(1+d)(1+5d),解得d=-2,所以S6=6×1+6×52×(-2)=-24,故选A.6.-12解析由已知得S1=a1,S2=a1+a2=2a1-1,S4=4a1+4×32×(-1)=4a1-6,而S1,S2,S4成等比数列,∴(2a1-1)2=a1(4a1-6),整理,得2a1+1=0,解得a1=-12.7.1解析设等差数列{an}的公差为d,等比数列{bn}的公比为q,由题意知-1+3d=-q3=8,即{-1+3d=8,-q3=8,解得{d=3,q=-2.故a2b2=-1+3-1×(-2)=1.8.4解析 a5-a1=15,a4-a2=6,∴{a1q4-a1=15,a1q3-a1q=6,(q≠1)两式相除得(q2+1)(q2-1)q(q2-1)=156,即2q2-5q+2=0,∴q=2或q=12,当q=2时,a1=1;当q=12时,a1=-16(舍去).∴a3=1×22=4.9.解(1)设等差数列{an}的公差为d,则依题意有{a1=1,(a1+3d)2=(a1+d)(a1+7d),解得d=1或d=0(舍去),∴an=1+(n-1)=n.(2)由(1)得an=n,∴bn=2n,∴bn+1bn=2,∴{bn}是首项为2,公比为2的等比数列,∴Tn=2(1-2n)1-2=2n+1-2.10.解(1)设等差数列{an}的公差为d. S4=4(a3+1),3a3=5a4,∴{4a1+6d=4(a1+2d+1),3a1+6d=5a1+15d,解得{a1=9,d=-2.∴an=11-2n.设数列{bn}的公比为q. b1b2=b3,2b1=a5,∴{b12q=b1q2,2b1=1,解得{b1=12,q=12.∴bn=(12)n.(2)由(1)知,Sn=10n-n2.由an=11-2n≤0可知n≥5.5,即a1>0,a2>0,…,a5>0,a6<0,a7<0,…,an<0.故当n≤5时,Tn=Sn=10n-n2;当n≥6时,Tn=2S5-Sn=n2-10n+50.于是Tn={10n-n2,n≤5,n2-10n+50,n≥6.11.解(1) S1=a1=1+ka1,∴a1=11-k.又an=Sn-Sn-1(n≥2),∴an=kk-1an-1(n≥2),∴an=11-k·(kk-1)n-1=-kn-1(k-1)n.(2) 在数列{an}中,a1=11-k,q=kk-1,∴{an2}是首项为(11-k)2,公比为(kk-1)2的等比数列.当k=-1时,等比数列{an2}的首项为14,公比为14...
