（天津专用）2020届高考数学一轮复习考点规范练24等比数列及其前n项和（含解析）新人教A版.docx

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(q0),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,an0,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(k0,且k1).(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d0,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b1 008=()A.2 017B.2 016C.2 015D.2 01414.设等比数列an满足a1+a3=10,a2+a4=5,则a1a2an的最大值为.15.在数列an中,a1=2,an+1=n+12nan(nN*).(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:Tn0,a20,a50,a60,a70,an0.故当n5时,Tn=Sn=10n-n2;当n6时,Tn=2S5-Sn=n2-10n+50.于是Tn=10n-n2,n5,n2-10n+50,n6.11.解(1)S1=a1=1+ka1,a1=11-k.又an=Sn-Sn-1(n2),an=kk-1an-1(n2),an=11-kkk-1n-1=-kn-1(k-1)n.(2)在数列an中,a1=11-k,q=kk-1,an2是首项为11-k2,公比为kk-12的等比数列.当k=-1时,等比数列an2的首项为14,公比为14,a12+a22+an2=141-14n1-14=131-14n.12.C解析因为ak是a6与ak+6的等比中项,所以ak2=a6ak+6.又等差数列an的公差d0,且a2=-d,所以a2+(k-2)d2=(a2+4d)a2+(k+4)d,所以(k-3)2=3(k+3),解得k=9或k=0(舍去),故选C.13.A解析在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a10,数列an为等比数列,an=84n-1=22n+1,bn=log2an=2n+1,b1008=2017,故选A.14.64解析由已知a1+a3=10,a2+a4=a1q+a3q=5,两式相除得a1+a3q(a1+a3)=105,解得q=12,a1=8,所以a1a2an=8n121+2+(n-1)=2-12n2+7n2,抛物线f(n)=-12n2+72n的对称轴为n=-722-12=3.5,又nN*,所以当n=3或n=4时,a1a2an取最大值为2-1232+732=26=64.15.证明(1)由题设得an+1n+1=12ann,又a11=2,所以数列ann是首项为2,公比为12的等比数列,所以ann=212n-1=22-n,an=n22-n=4n2n.(2)由(1)知bn=an4n-an=4n2n4n-4n2n=12n-1,因为对任意nN*,2n-12n-1,所以bn12n-1.所以Tn1+12+122+123+12n-1=21-12n2.16.(1)证明an+1=an+6an-1(n2),an+1+2an=3an+6an-1=3(an+2an-1)(n2).又a1=5,a2=5,a2+2a1=15,an+2an-10(n2),an+1+2anan+2an-1=3(n2),数列an+1+2an是以15为首项,3为公比的等比数列.(2)解由(1)得an+1+2an=153n-1=53n,则