课时规范练29 等比数列及其前n项和基础巩固组1.已知等比数列{an}满足a1=14,a3a5=4(a4-1),则a2=( ) A.2B.1C.12D.182.在正项等比数列{an}中,a2,a48是方程2x2-7x+6=0的两个根,则a1·a2·a25·a48·a49的值为( )A.212B.93C.±93D.353.(2022安徽黄山市二模)已知数列{an}的前n项和为Sn,且a1=2,an+1=Sn+1(n∈N*),则S5=( )A.31B.42C.37D.474.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=3-2an5.(2022全国Ⅲ)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.设等比数列{an}的前n项和为Sn.若S2=3,S4=15,则S6=( )A.31B.32C.63D.647.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 . 8.(2022北京)若等差数列{an}和等比数列{bn}满足a1=b1=-1,a4=b4=8,则a2b2= . 9.(2022江苏,9)等比数列{an}的各项均为实数,其前n项和为Sn.已知S3=74,S6=634,则a8= . 10.(2022河南新乡二模,文17)在数列{an}中,a1=12,{an}的前n项和Sn满足Sn+1-Sn=12n+1(n∈N*).(1)求数列{an}的通项公式an以及前n项和Sn;(2)若S1+S2,S1+S3,m(S2+S3)成等差数列,求实数m的值.5\n〚导学号24190754〛综合提升组11.(2022四川广元二诊)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.201412.设等比数列{an}满足a1+a3=10,a2+a4=5,则a1·a2·a3·…·an的最大值为 . 13.已知{an}是公差为3的等差数列,数列{bn}满足b1=1,b2=13,anbn+1+bn+1=nbn.(1)求{an}的通项公式;(2)求{bn}的前n项和.创新应用组14.已知数列{an}的前n项和为Sn,在数列{bn}中,b1=a1,bn=an-an-1(n≥2),且an+Sn=n.(1)设cn=an-1,求证:{cn}是等比数列;(2)求数列{bn}的通项公式.答案：5\n1.C ∵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 ∵a2,a48是方程2x2-7x+6=0的两个根,∴a2·a48=3.又a1·a49=a2·a48=a252=3,a25>0,∴a1·a2·a25·a48·a49=a255=93.3.D ∵an+1=Sn+1(n∈N*),∴Sn+1-Sn=Sn+1(n∈N*),∴Sn+1+1=2(Sn+1)(n∈N*),∴数列{Sn+1}是首项为3,公比为2的等比数列.则S5+1=3×24,解得S5=47.4.D Sn=a1(1-qn)1-q=a1-anq1-q=1-23an1-23=3-2an,故选D.5.A 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.C ∵S2=3,S4=15,∴由等比数列前n项和的性质,得S2,S4-S2,S6-S4成等比数列,∴(S4-S2)2=S2(S6-S4),即(15-3)2=3(S6-15),解得S6=63,故选C.7.-12 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=-12.8.1 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)=1.9.32 设该等比数列的公比为q,则S6-S3=634-74=14,即a4+a5+a6=14.①∵S3=74,∴a1+a2+a3=74.由①得(a1+a2+a3)q3=14,∴q3=1474=8,即q=2.5\n∴a1+2a1+4a1=74,a1=14,∴a8=a1·q7=14×27=32.10.解(1)∵an+1=Sn+1-Sn=12n+1,∴当n≥2时,an=12n.又a1=12,∴当n=1时上式也成立.∴an=12n,∴Sn=121-12n1-12=1-12n.(2)由(1)可得:S1=12,S2=34,S3=78.∵S1+S2,S1+S3,m(S2+S3)成等差数列,∴12+34+m34+78=212+78,解得m=1213.11.A 在an=34Sn+2中,令n=1得a1=8,∵an=34Sn+2成立,∴an+1=34Sn+1+2成立,两式相减得an+1-an=34an+1,∴an+1=4an,又a1≠0,∴数列{an}为等比数列,∴an=8·4n-1=22n+1,∴bn=log2an=2n+1,∴b1008=2017,故选A.12.64 由已知a1+a3=10,a2+a4=(a1+a3)q=5,得q=12,所以a1=8,所以a1·a2·a3·…·an=8n·121+2+…+(n-1)=2-12n2+7n2,所以当n=3或n=4时,a1·a2·a3·…·an取最大值为2-12×32+7×32=26=64.13.解(1)由已知,得a1b2+b2=b1,因为b1=1,b2=13,所以a1=2.所以数列{an}是首项为2,公差为3的等差数列,通项公式为an=3n-1.(2)由(1)和anbn+1+bn+1=nbn,得bn+1=bn3,因此{bn}是首项为1,公比为13的等比数列.记{bn}的前n项和为Sn,则Sn=1-13n1-13=32-12×3n-1.5\n14.(1)证明∵an+Sn=n,①∴an+1+Sn+1=n+1.②②-①得an+1-an+an+1=1,∴2an+1=an+1,∴2(an+1-1)=an-1,∴an+1-1an-1=12,∴{an-1}是等比数列.又a1+a1=1,∴a1=12,∵首项c1=a1-1,∴c1=-12,公比q=12.又cn=an-1,∴{cn}是以-12为首项,以12为公比的等比数列.(2)解由(1)可知cn=-12·12n-1=-12n,∴an=cn+1=1-12n.∴当n≥2时,bn=an-an-1=1-12n-1-12n-1=12n-1-12n=12n.又b1=a1=12代入上式也符合,∴bn=12n.5
