2024高考總復(fù)習(xí)優(yōu)化設(shè)計一輪用書數(shù)學(xué)配北師版(適用于老高考新教材)課時規(guī)范練27　等比數(shù)列

課時規(guī)范練27等比數(shù)列基礎(chǔ)鞏固組1.(2021山東泰安高三月考)已知各項均為正數(shù)的等比數(shù)列{an}的前n項和為Sn,若S4-S3=27,S3-S1=12,則S5=()A.81 B.812 C.121 D.2.(2021遼寧沈陽高三期中)在等比數(shù)列{an}中,a1+a2+a3=3,a4+a5+a6=6,則數(shù)列{an}的前12項和為()A.90 B.60 C.45 D.323.(2021陜西高新一中高三二模)已知數(shù)列{an}是等比數(shù)列,Sn是其前n項積,若S7S2=32,則S9=A.1024 B.512 C.256 D.1284.(2021云南曲靖高三月考)設(shè)等比數(shù)列{an}的前n項和為Sn,若S3,S9,S6成等差數(shù)列,且a8=6,則a11=()A.-1 B.-3 C.-5 D.-75.(2021湖南長沙高三期末)在公比不為1的等比數(shù)列{an}中,存在s,t∈N*,滿足asat=a52,則4s+1A.34 B.7C.58 D.6.在各項均為正數(shù)的等比數(shù)列{an}中,Sn是數(shù)列{an}的前n項和,若a1=12,S3=78,則下列說法錯誤的是(A.an≤12B.an≥12C.Sn<1 D.2lgan=lgan-2+lgan+2(n≥3)7.已知等比數(shù)列{an}的公比為2,且S1,S2+2,S3成等差數(shù)列,則下列說法錯誤的是()A.an=2nB.a2,a3,a4-4成等差數(shù)列C.{Sn+2}是等比數(shù)列D.?m,n,r∈N*(m≠r),am,an,ar成等差數(shù)列8.在數(shù)列{an}中,a1≠0,an+1=2an,Sn為數(shù)列{an}的前n項和,則S6-S39.(2021北京石景山高三月考)已知Sn是等比數(shù)列{an}的前n項和,|an+1|>|an|,a2=-2,S3=3,則|a1|+|a2|+…+|an|=.

綜合提升組10.(2021江蘇泰州高三月考)已知λ為非零常數(shù),數(shù)列{an}與{2an+λ}均為等比數(shù)列,且a2021=3,則a1=()A.1 B.3 C.2021 D.404211.已知數(shù)列{an}的前n項和為Sn,若a1+a2=2,an+1=Sn+1,則()A.數(shù)列{an}是公比為2的等比數(shù)列B.S6=49C.anD.1a1+112.(2021福建師大附中高三模擬)已知數(shù)列{an}為等比數(shù)列,若數(shù)列{3n-an}也是等比數(shù)列,則數(shù)列{an}的通項公式可以為.(寫出一個即可)

13.(2021山東煙臺高三期中)在等比數(shù)列{an}中,a1=1,q=12,記Tn=a22+a42+a6214.(2021天津耀華中學(xué)高三月考)在數(shù)列{an}中,a1=1,a2=3,an+2=3an+1-2an.(1)證明:數(shù)列{an+1-an}是等比數(shù)列;(2)求數(shù)列{an}的通項公式.創(chuàng)新應(yīng)用組15.(2021浙江杭州高三模擬)已知公比不為1的各項均為正數(shù)的等比數(shù)列{an}的前n項和為Sn,數(shù)列{bn}滿足bn=S2nSA.7b2≤8b6,b2≥b4+1B.7b2≤8b6,b2≤b4+1C.3b3≤4b9,b4≥b8+1D.3b3≤4b9,b4≤b8+116.已知數(shù)列{an}的前4項成等比數(shù)列,其前n項和為Sn,且S4=lnS3,a1>1,則()A.a1<a3 B.a1<a2C.a1>a3 D.a2<a4

課時規(guī)范練27等比數(shù)列1.C解析:設(shè)等比數(shù)列{an}的公比為q(q>0).由題意S4-S3=27,S3-S1=12,可得a4=27,a2+a3=12,所以a1q3=27,a1q+a1q22.C解析:設(shè)數(shù)列{an}的公比為q,則a4+a5+a6a1+a2+a3=q3=63=2,所以a7+a8+a9=(a4+a5+a6)q3=6×2=12,同理a10+a11+a12=24,所以S12=a1+a2+…3.B解析:因為S7S2=a3a4a5a6a7=(a5)5=32,所以a5=2,所以S9=a1a2a3a4a5a6a7a8a9=(a5)9=512.4.B解析:設(shè)等比數(shù)列{an}的公比為q,由S3,S9,S6成等差數(shù)列,可得2S9=S3+S6,即2(S9-S6)=S3-S6,即2(a7+a8+a9)=-(a4+a5+a6),即q3=-12.又a8=6,所以a11=a8q3=6×-12=-3.故選B.5.C解析:設(shè)等比數(shù)列{an}的公比為q(q≠1).因為asat=a52,所以a1qs-1·a1qt-1=(a1q4)2,即qs+t-2=q8,可得s+t=10,且s,t∈N*,則4s+14t=110·4s+14t(s+t)=1104+4ts+s4t+14=110174+4ts+s4t.因為s,t∈N*,所以4t6.B解析:設(shè)等比數(shù)列{an}的公比為q(q>0).由a1=12,S3=78,得12+12q+12q2=78,即4q2+4q-3=0,解得q=12或q=-32(舍去),所以an=12×12n-1=12n.又n∈N*,所以an≤12,故選項A正確,選項B錯誤;Sn=12[1-(12)

n]1-12=1-12n<1,故選項C正確;2lgan=2lg12n=lg122n=lg12n-2+lg12n+2=7.D解析:由S1,S2+2,S3成等差數(shù)列,可得a2=4,a1=2,an=2n,故選項A正確;a3=8,a4-4=12,a2,a3,a4-4成等差數(shù)列,故選項B正確;Sn=2n+1-2,所以Sn+2=2n+1,所以{Sn+2}是等比數(shù)列,故選項C正確;若am,an,ar即2m,2n,2r成等差數(shù)列,不妨設(shè)m<n<r,則2m+2r=2·2n,2m(1+2r-m)=2n+1-m·2m,即1+2r-m=2n+1-m,顯然左邊為奇數(shù),右邊為偶數(shù),不相等,故選項D錯誤.故選D.8.8解析:∵a1≠0,an+1=2an,∴an+1an=2=q(q為公比),∴S69.2n-1解析:設(shè)等比數(shù)列{an}的公比為q,易知q≠1.因為a2=-2,S3=3,所以a1q因為|an+1|>|an|,所以|q|>1,所以數(shù)列{an}是以1為首項,以-2為公比的等比數(shù)列,則{|an|}也為等比數(shù)列,公比為2,首項|a1|=1,故|a1|+|a2|+…+|an|=1-2n1-210.B解析:{an}與{2an+λ}均為等比數(shù)列,所以(2an+λ)2=(2an-1+λ)(2an+1+λ)且an2=an-1an+1,得2an=an-1+an+1,故數(shù)列{an}也為等差數(shù)列,所以數(shù)列{an}為非零常數(shù)列,則a1=a2021=3.故選11.D解析:an+1=Sn+1,令n=1,得a2=S1+1=a1+1,結(jié)合a1+a2=2,知a1=12,a2=32.又an+1=Sn+1,所以an=Sn-1+1(n≥2),所以an+1=2an,但a1=12,a2=32,當(dāng)n≥2時,Sn=32(1-2n-1)1-2+12=3×2n-2-1,S6=3×16-1=47,故選項A,B錯誤;因為a1S1=1,當(dāng)n≥2時,anSn=3×2n-33×2n12.an=3n-1(答案不唯一)解析:設(shè)數(shù)列{an}的公比為q,∵數(shù)列{3n-an}是等比數(shù)列,∴(32-a1q)2=(3-a1)(33-a1q2),即q2-6q+9=0,解得q=3.取a1=1,則an=3n-1.答案不唯一.13.4151-124n解析:由題知,an=1×12n-1=12n-1,則a2n2=122n-12=424n,則Tn=a14.(1)證明因為an+2=3an+1-2an,所以an+2-又因為a1=1,a2=3,所以a2-a1=2,則數(shù)列{an+1-an}是以2為首項,2為公比的等比數(shù)列.(2)解由(1)知an+1-an=2×2n-1=2n,所以an-an-1=2n-1,an-1-an-2=2n-2,…,a3-a2=22,a2-a1=21,則(an-an-1)+(an-1-an-2)+…+(a3-a2)+(a2-a1)=2n-1+2n-2+…+22+21,即an-a1=2×(1-2n-1)115.D解析:設(shè)數(shù)列{an}的公比為q(q>0,q≠1),則Sn=a1(1-qn)1-q,S2n=a1(1-q2n)1-q,∴bn=S2nSn=1-q2n1-qn=(1+qn)(1-qn)1-qn=1+q∵b9b3?34=1+q91+q3?34=(1+q3)(1-q3+q6)1+q3?34=q6-q3+14=q3-122>0,即b9b3≥34,∴3b3≤4b9.又b4-b8-116.A解析:∵數(shù)列{an}的前4項成等比數(shù)列,且S4=lnS