高考數(shù)學(xué)（理科課標(biāo)版）一輪復(fù)習(xí)題組訓(xùn)練第6章第3講等比數(shù)列及其前n項(xiàng)和

第三講等比數(shù)列及其前n項(xiàng)和題組1等比數(shù)列及其前n項(xiàng)和1.[2015新課標(biāo)全國Ⅱ,4,5分][理]已知等比數(shù)列{an}滿足a1=3,a1+a3+a5=21,則a3+a5+a7=()A.21 B.42 C.63 D.842.[2013新課標(biāo)全國Ⅱ,3,5分][理]等比數(shù)列{an}的前n項(xiàng)和為Sn.已知S3=a2+10a1,a5=9,則a1=()A.13 B.13 C.193.[2017江蘇,9,5分][理]等比數(shù)列{an}的各項(xiàng)均為實(shí)數(shù),其前n項(xiàng)和為Sn.已知S3=74,S6=634,則a8=4.[2015新課標(biāo)全國Ⅰ,13,5分]在數(shù)列{an}中,a1=2,an+1=2an,Sn為{an}的前n項(xiàng)和.若Sn=126,則n=.

5.[2013遼寧,14,5分][理]已知等比數(shù)列{an}是遞增數(shù)列,Sn是{an}的前n項(xiàng)和.若a1,a3是方程x25x+4=0的兩個(gè)根,則S6=.

6.[2017山東,19,12分][理]已知{xn}是各項(xiàng)均為正數(shù)的等比數(shù)列,且x1+x2=3,x3x2=2.(Ⅰ)求數(shù)列{xn}的通項(xiàng)公式;(Ⅱ)如圖631,在平面直角坐標(biāo)系xOy中,依次連接點(diǎn)P1(x1,1),P2(x2,2),…,Pn+1(xn+1,n+1)得到折線P1P2…Pn+1,求由該折線與直線y=0,x=x1,x=xn+1所圍成的區(qū)域的面積Tn.圖6317.[2016全國卷Ⅲ,17,12分][理]已知數(shù)列{an}的前n項(xiàng)和Sn=1+λan,其中λ≠0.(Ⅰ)證明{an}是等比數(shù)列,并求其通項(xiàng)公式;(Ⅱ)若S5=3132,求題組2等比數(shù)列的性質(zhì)8.[2014重慶,2,5分][理]對(duì)任意等比數(shù)列{an},下列說法一定正確的是()A.a1,a3,a9成等比數(shù)列B.a2,a3,a6成等比數(shù)列C.a2,a4,a8成等比數(shù)列D.a3,a6,a9成等比數(shù)列9.[2014廣東,13,5分][理]若等比數(shù)列{an}的各項(xiàng)均為正數(shù),且a10a11+a9a12=2e5,則lna1+lna2+…+lna20=.

10.[2016天津,18,13分]已知{an}是等比數(shù)列,前n項(xiàng)和為Sn(n∈N*),且1a11a2=2a3(Ⅰ)求{an}的通項(xiàng)公式;(Ⅱ)若對(duì)任意的n∈N*,bn是log2an和log2an+1的等差中項(xiàng),求數(shù)列{(1)nbn2}的前2nA組基礎(chǔ)題1.[2018河北衡水中學(xué)二調(diào),3]設(shè)正項(xiàng)等比數(shù)列{an}的前n項(xiàng)和為Sn,且an+1an<1,若a3+a5=20,a3a5=64,則S4=A.63或120 B.256 C.120 D.632.[2018益陽市、湘潭市高三調(diào)研,4]已知等比數(shù)列{an}中,a5=3,a4a7=45,則a7-a9aA.3 B.5 C.9 D.253.[2018洛陽市尖子生高三第一次聯(lián)考,7]在等比數(shù)列{an}中,a3,a15是方程x2+6x+2=0的根,則a2a16a9A.2+22 B.2C.2 D.24.[2017吉林省部分學(xué)校高三仿真考試,7][數(shù)學(xué)文化題]《張丘建算經(jīng)》中“今有馬行轉(zhuǎn)遲,次日減半,疾七日,行七百里.問日行幾何?”意思是:“現(xiàn)有一匹馬行走的速度逐漸變慢,每天走的里數(shù)是前一天的一半,連續(xù)行走7天,共走了700里路,問每天走的里數(shù)為多少?”則該匹馬第一天走的里數(shù)為()A.128127 B.44800127 C.5.[2018廣州市高三調(diào)研測(cè)試,14]在各項(xiàng)都為正數(shù)的等比數(shù)列{an}中,若a2018=22,則1a20176.[2018惠州市一調(diào),15]已知等比數(shù)列{an}的公比為正數(shù),且a3a9=2a52,a2=1,則a1=7.[2017昆明市高三適應(yīng)性檢測(cè),17]數(shù)列{an}滿足a1=1,an+1+2an=3.(1)證明{an1}是等比數(shù)列,并求數(shù)列{an}的通項(xiàng)公式;(2)已知符號(hào)函數(shù)sgn(x)=1,x>0,0,x=0,-1,xB組提升題8.[2018石家莊市重點(diǎn)高中摸底考試,9]已知等比數(shù)列{an}的前n項(xiàng)和為Sn,若S3=7,S6=63,則數(shù)列{nan}的前n項(xiàng)和為()A.3+(n+1)×2n B.3+(n+1)×2nC.1+(n+1)×2n D.1+(n1)×2n9.[2017重慶市名校聯(lián)盟二診,10分]設(shè)Tn為等比數(shù)列{an}的前n項(xiàng)之積,且a1=6,a4=34,則當(dāng)Tn最大時(shí),n的值為()A.4 B.6 C.8 D.1010.[2017天星第二次大聯(lián)考,7]已知數(shù)列{an}的前n項(xiàng)和為Sn,若4nSn(6n3)an=3n,則下列說法正確的是()A.數(shù)列{an}是以3為首項(xiàng)的等比數(shù)列B.數(shù)列{an}的通項(xiàng)公式為an=3C.數(shù)列{anD.數(shù)列{ann11.[2017鄭州市第三次質(zhì)量預(yù)測(cè),8]已知等比數(shù)列{an},且a6+a8=0416-x2dx,則a8(a4+2a6+aA.π2 B.4π2 C.8π2 D.16π212.[2017太原市高三三模,9]已知數(shù)列{an}的前n項(xiàng)和為Sn,點(diǎn)(n,Sn+3)(n∈N*)在函數(shù)y=3×2x的圖象上,等比數(shù)列{bn}滿足bn+bn+1=an(n∈N*),其前n項(xiàng)和為Tn,則下列結(jié)論正確的是()A.Sn=2Tn B.Tn=2bn+1C.Tn>an D.Tn<bn+113.[2018河北省“五個(gè)一名校聯(lián)盟”高三第二次考試,17]已知數(shù)列{an}是等差數(shù)列,a2=6,前n項(xiàng)和為Sn,數(shù)列{bn}是等比數(shù)列,b2=2,a1b3=12,S3+b1=19.(1)求{an},{bn}的通項(xiàng)公式;(2)求數(shù)列{bncos(anπ)}的前n項(xiàng)和Tn.答案1.B設(shè)等比數(shù)列{an}的公比為q,則a1(1+q2+q4)=21,又a1=3,所以q4+q26=0,所以q2=2(q2=3舍去),所以a3=6,a5=12,a7=24,所以a3+a5+a7=42.故選B.2.C由題知q≠1,所以S3=a1(1-q3)1-q=a1q+10a1,解得q2=9,又a5=a13.32設(shè)等比數(shù)列{an}的公比為q,由S6≠2S3得q≠1,則S3=a1(1-q3)1-q=74,S6=a1(1-q6)1-q=6344.6因?yàn)閍1=2,an+1=2an,所以數(shù)列{an}是首項(xiàng)為2,公比為2的等比數(shù)列,因?yàn)镾n=126,所以2-2n+11-2=126,解得25.63因?yàn)閍1,a3是方程x25x+4=0的兩個(gè)根,且數(shù)列是遞增數(shù)列,所以a1=1,a3=4,所以q=2,代入等比數(shù)列的求和公式得S6=1-266.(Ⅰ)設(shè)數(shù)列{xn}的公比為q,由已知知q>0.由題意得x所以3q25q2=0.因?yàn)閝>0,所以q=2,x1=1,因此數(shù)列{xn}的通項(xiàng)公式為xn=2n1.(Ⅱ)過P1,P2,…,Pn+1向x軸作垂線,垂足分別為Q1,Q2,…,Qn+1.由(Ⅰ)得xn+1xn=2n2n1=2n1,記梯形PnPn+1Qn+1Qn的面積為bn,由題意得bn=(n+n+1)2×2n1=(2n+所以Tn=b1+b2+…+bn=3×21+5×20+7×21+…+(2n1)×2n3+(2n+1)×2n2①,則2Tn=3×20+5×21+7×22+…+(2n1)×2n2+(2n+1)×2n1②,①②,得Tn=3×21+(2+22+…+2n1)(2n+1)×2n1=32+2(1-2n-1所以Tn=(27.(Ⅰ)由題意得a1=S1=1+λa1,故λ≠1,a1=11-λ,a由Sn=1+λan,Sn+1=1+λan+1得an+1=λan+1λan,即an+1(λ1)=λan.由a1≠0,λ≠0且λ≠1得an≠0,所以an+1a所以{an}是首項(xiàng)為11-λ故an=11-λ(λλ(Ⅱ)由(Ⅰ)得Sn=1(λλ-1)n.由S5=3132得1(λλ-1)5=3132,即(λ8.D由等比數(shù)列的性質(zhì),得a3·a9=a62≠0,因此a3,a6,a99.50由等比數(shù)列的性質(zhì)可知a1a20=a2a19=…=a9a12=a10a11=e5,于是a1a2…a20=(e5)10=e50,lna1+lna2+…+lna20=ln(a1a2…a20)=lne50=50.10.(Ⅰ)設(shè)數(shù)列{an}的公比為q.由已知,得1a11a1q=2a又由S6=a1·1-q61-所以a1·1-261-2=63,得a1=1.所以a(Ⅱ)由題意,得bn=12(log2an+log2an+1)=12(log22n1+log22n)=n12,即{bn}是首項(xiàng)為設(shè)數(shù)列{(1)nbn2}的前n項(xiàng)和為TT2n=(b12+b22)+(b32+b42)=b1+b2+b3+b4+…+b2n1+b2n=2=2n2.A組基礎(chǔ)題1.C由題意得a3+a5=20,a3a5=64,解得a3=16,a5=4或a3=4,a5=16.又an+1an<1,所以數(shù)列{an}為遞減數(shù)列,故a2.D設(shè)等比數(shù)列{an}的公比為q,則a4a7=a5q·a5q2=9所以q=5,所以a7-a9a5-a73.B設(shè)等比數(shù)列{an}的公比為q,因?yàn)閍3,a15是方程x2+6x+2=0的根,所以a3·a15=a92=2,a3+a15=6,所以a3<0,a15<0,則a9=2,所以a2a16a9=4.B由題意知該匹馬每日所走的路程成等比數(shù)列{an},且公比q=12,S7=700,由等比數(shù)列的求和公式得Sn=a1(1-1275.4設(shè)公比為q(q>0),因?yàn)閍2018=22,所以a2017=a2018q=22q,a2019=a2018q=22q,則有1a2017+2a2019=2q+2226.22∵a3a9=a62,∴a62=2a52,設(shè)等比數(shù)列{an}的公比為q,∴q2=2,由于q>0,解得q=2,∴a17.(1)因?yàn)閍n+1=2an+3,a1=1,所以an+11=2(an1),a11=2,所以數(shù)列{an1}是首項(xiàng)為2,公比為2的等比數(shù)列.故an1=(2)n,即an=(2)n+1.(2)bn=an·sgn(an)=2設(shè)數(shù)列{bn}的前n項(xiàng)和為Sn,則S100=(21)+(22+1)+(231)+…+(2991)+(2100+1)=2+22+23+…+2100=21012.B組提升題8.D設(shè)等比數(shù)列{an}的公比為q,∵S3=7,S6=63,∴q≠1,∴a1(1-q3)1-q=7,a1(1-q6)1-q=63,解得a1=1,q=2,∴an=2n1,∴nan=n·2n1,設(shè)數(shù)列{nan}的前2Tn=2+2·22+3·23+4·24+…+(n1)·2n1+n·2n,兩式相減得Tn=1+2+22+23+…+2n1n·2n=2n1n·2n=(1n)2n1,∴Tn=1+(n1)×2n,故選D.9.A設(shè)等比數(shù)列{an}的公比為q,∵a1=6,a4=34,∴34=6q3,解得q=12,∴an=6×(12∴Tn=(6)n×(12)0+1+2+…+(n1)=(6)n×(12)n(n-1)2,當(dāng)n為奇數(shù)時(shí),Tn<0,當(dāng)n為偶數(shù)時(shí),Tn>0,故當(dāng)n為偶數(shù)時(shí),Tn才有可能取得最大值.∵T2k=36k×(12)k(2k1),∴T2k+2T2k=36k+1×(1∴T2<T4,T4>T6>T8>…,則當(dāng)Tn最大時(shí),n的值為4.10.C解法一由4nSn(6n3)an=3n,可得4Sn=3n+(6n-3)ann當(dāng)n=1時(shí),4S1(63)a1=3,解得a1=3,當(dāng)n≥2時(shí),4Sn1=3+(63n-1)an1由①②得,4an=(63n)an(63n-1)an1,整理得2n-3nan=6n-9n-解法二當(dāng)n=1時(shí),4S1(63)a1=3,解得a1=3,當(dāng)n=2時(shí),8(a1+a2)(6×23)a2=3×2,解得a2=18,當(dāng)n=3時(shí),12(a1+a2+a3)(6×33)a3=3×3,解得a3=81,B錯(cuò)誤;又a2a1=183=6,a3a2=8118=92,故A錯(cuò)誤;a11=3,11.D因?yàn)閍6+a8=0416-x2dx=14×π×42=4π,所以a8(a4+2a6+a8)=a8a4+2a6a8+a82=a62+2a612.D因?yàn)辄c(diǎn)(n,Sn+3)(n∈N*)在函數(shù)y=3×2x的圖象上,所以Sn=3·2n3,所以Sn1=3·2n13,兩式相減得an=3·2n1,所以bn+bn+1=3·2n1,因?yàn)閿?shù)列{bn}為等比數(shù)列,設(shè)公比為q,則b1+b1q=3,b2+b2q=6,解得b1=1,q=2