全國(guó)統(tǒng)考高考數(shù)學(xué)大一輪復(fù)習(xí)第6章數(shù)列第3講等比數(shù)列及其前n項(xiàng)和2備考試題文含解析

一輪復(fù)習(xí)精品資料(高中)PAGEPAGE1第六章數(shù)列第三講等比數(shù)列及其前n項(xiàng)和1.〖2021陜西百校聯(lián)考〗已知等比數(shù)列{an}的公比為q,前4項(xiàng)的和為a1+14,且a2,a3+1,a4成等差數(shù)列,則q的值()A.12或2 B.1或122.〖2021安徽省四校聯(lián)考〗已知正項(xiàng)等比數(shù)列{an}的前n項(xiàng)和為Sn,若a4=18,S3-a1=34,則S4=(A.116 B.18 C.313.〖2020合肥三檢〗〖數(shù)學(xué)文化題〗公元前1650年左右的埃及《萊因德紙草書(shū)》上載有如下問(wèn)題:“十人分十斗玉米,從第二人開(kāi)始,各人所得依次比前人少八分之一,問(wèn)每人各得玉米多少斗?”在上述問(wèn)題中,第一人分得玉米()A.70×89810-1斗 B.10×81084.〖2020南昌市測(cè)試〗公比不為1的等比數(shù)列{an}中,若a1a5=aman,則mn不可能為()A.5 B.6 C.8 D.95.〖2020成都市高三摸底測(cè)試〗已知等比數(shù)列{an}的各項(xiàng)均為正數(shù),若log3a1+log3a2+…+log3a12=12,則a6a7=()A.1 B.3 C.6 D.96.〖2021四省八校聯(lián)考〗已知等比數(shù)列{an}的公比為q,前n項(xiàng)和為Sn=m-qn,若a5=-8a2,則S5=.

7.〖2020大同市高三調(diào)研〗已知各項(xiàng)均為正數(shù)的等比數(shù)列{an}中,a1a2a3=5,a7a8a9=10,則a4a5a6=.

8.〖2020全國(guó)卷Ⅲ,17,12分〗設(shè)等比數(shù)列{an}滿足a1+a2=4,a3-a1=8.(1)求{an}的通項(xiàng)公式;(2)記Sn為數(shù)列{log3an}的前n項(xiàng)和.若Sm+Sm+1=Sm+3,求m.9.〖條件創(chuàng)新〗已知等比數(shù)列{an}的前n項(xiàng)和為Sn,且an+1+λ3=Sn,a3=12,則實(shí)數(shù)λA.-34 B.-14 C.10.〖設(shè)問(wèn)創(chuàng)新〗已知等比數(shù)列{an}的前5項(xiàng)積為32,1<a1<2,則a1+a32+a5A.〖3,72) B.(3,+∞) C.(3,72) D.11.〖2020成都市三診〗在等比數(shù)列{an}中,已知anan+1=9n,則該數(shù)列的公比是()A.-3 B.3 C.±3 D.912.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.1013.〖2021四省八校聯(lián)考〗設(shè)無(wú)窮數(shù)列{an}的前n項(xiàng)和為Sn,有三個(gè)條件:①am+n=am·an,②Sn=an+1+1,a1≠0,③Sn=2an+1p(p是與n無(wú)關(guān)的參數(shù)),則從中選出兩個(gè)條件,能使數(shù)列{an}為唯一確定的等比數(shù)列的條件是14.〖2020安徽省示范高中名校聯(lián)考〗設(shè)Sn是各項(xiàng)均為正數(shù)的等比數(shù)列{an}的前n項(xiàng)和,a1=3,若-a4,a3,a5成等差數(shù)列,則Sn與an的關(guān)系式為.

15.已知公比q>1的等比數(shù)列{an}滿足a52=a10,2(an+an+2)=5an+1.若bn=(n-λ)an(n∈N*),且數(shù)列{bn}是遞增數(shù)列,則實(shí)數(shù)λ的取值范圍是16.〖2020海南,18,12分〗已知公比大于1的等比數(shù)列{an}滿足a2+a4=20,a3=8.(1)求{an}的通項(xiàng)公式;(2)求a1a2-a2a3+…+(-1)n-1anan+1.答案第六章數(shù)列第三講等比數(shù)列及其前n項(xiàng)和1.A由題意知a1+a2+a3+a4=a1+14,且a2+a4=2(a3+1),兩式聯(lián)立,可解得a3=4,所以a2+a4=10,a2a4=a32=16,解得a2=2,a4=8或a2.D解法一設(shè)等比數(shù)列{an}的公比為q(q>0且q≠1),∵a4=18,S3-a1=34,∴a1(1-q3)1-q-解法二設(shè)等比數(shù)列{an}的公比為q(q>0且q≠1),∵S3-a1=a2+a3=a4q2+a4q=34,a4=18,∴q=12,∴a1=a4q3=1,S43.C設(shè)第i個(gè)人分到的玉米斗數(shù)為ai(i=1,2,…,9,10),則{an}是公比為78的等比數(shù)列.由題意知a1〖1-(78)4.B由等比數(shù)列的性質(zhì)可知,m+n=6,m∈N*,n∈N*,當(dāng)m=n=3時(shí),mn=9;當(dāng)m=4,n=2時(shí),mn=8;當(dāng)m=5,n=1時(shí),mn=5.故選B.5.D因?yàn)榈缺葦?shù)列{an}的各項(xiàng)均為正數(shù),所以log3a1+log3a2+…+log3a12=log3(a1·a2·…·a12)=log3(a6a7)6=12,所以(a6a7)6=312=96,所以a6a7=9,故選D.6.33由a5=-8a2得a1q4=-8a1q(a1≠0),解得q=-2,則Sn=a1(1-qn)1-q=a13-a137.52各項(xiàng)均為正數(shù)的等比數(shù)列{an}中,a1a2a3=a23=5,a7a8a9=a83=10,則a4a5a6=8.(1)設(shè){an}的公比為q,則an=a1qn-1.由已知得a1+a1q=4,所以{an}的通項(xiàng)公式為an=3n-1.(2)由(1)知log3an=n-1.故Sn=n(由Sm+Sm+1=Sm+3得m(m-1)+(m+1)m=(m+3)(m+2),即m2-5m-6=0.解得m=-1(舍去)或m=6.9.A由條件得an+1+λ=3Sn,當(dāng)n≥2時(shí),an+λ=3Sn-1,兩式相減,得an+1=4an,又a3=12,所以a2=3,a1=34,將n=1代入an+1+λ=3Sn,得a2+λ=3a1,得λ=-3410.C因?yàn)榈缺葦?shù)列{an}的前5項(xiàng)積為32,所以a35=32,解得a3=2,則a5=a32a1=4a1,a1+a32+a54=a1+1+111.B設(shè){an}的公比為q,根據(jù)題意知a1a2=9,a2a3=92,所以9=a2a3a1a2=a3a1=q2?q=±3.又a112.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)n-1.∴Tn=(-6)n×(12)0+1+2+…+(n-1)=(-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(2k-1),∴T2k+2T2k=36k+1×(12)(k+1)(2k13.①③在①中,令m=n=1,得a2=a12;在②中,Sn=an+1+1,當(dāng)n≥2時(shí),Sn-1=an+1,兩式相減,得an=an+1-an,即an+1=2an;在③中,Sn=2an+1p,Sn+1=2an+1+1p,兩式相減,得an+1=2an+1-2an,即an+1=2an.若選①②,則a2=a12,a1=a2+1,即a12=a1-1,a12-a1+1=0,Δ=(-1)2-4×1×1=-3<0,方程無(wú)解,故不能選①②作為條件;若選①③,則由an+1=214.Sn=2an-3設(shè)等比數(shù)列{an}的公比為q,因?yàn)閿?shù)列{an}的各項(xiàng)均為正數(shù),所以q>0.由-a4,a3,a5成等差數(shù)列,得2a3=a5-a4,則q2-q-2=0,解得q=2,所以Sn=a1(1-qn)1-q=a1-15.(-∞,3)2(an+an+2)=5an+1?2q2-5q+2=0?q=2或q=12(舍去),a52=a10?(a1q4)2=a1q9?a1=q=2,所以數(shù)列{an}的通項(xiàng)公式為an=2n,所以bn=(n-λ)2n(n∈N*),所以bn+1=(n+1-λ)·2n+1.因?yàn)閿?shù)列{bn}是遞增數(shù)列,所以bn+1>bn,所以(n+1-λ)2n+1>(n-λ)2n,化簡(jiǎn)得λ<n+2.因?yàn)閚∈N*,所以〖核心素養(yǎng)〗試題設(shè)計(jì)步步緊扣,求等比數(shù)列通項(xiàng)公式的過(guò)程體現(xiàn)了數(shù)學(xué)運(yùn)算核心素養(yǎng),利用遞增數(shù)列解題的過(guò)程則體現(xiàn)了邏輯推理核心素養(yǎng).16.(1)設(shè){an}的公比為q.由題設(shè)得a1q+a1q3=20,a1q2=8.解得q=12(舍去),q=2.由題設(shè)得a1=2所以{an}的通項(xiàng)公式為an=2n.(2)由(1)可知an=2n,則(-1)n-1anan+1=(-1)n-1×2n×2n+1=8×(-4)n-1,記Tn=a1a2-a2a3+…+(-1)n-1anan+1,則Tn=8×(-4)0+8×(-4)1+…+8×(-4)n-1=8×1=85〖1-(-4)n〗第六章數(shù)列第三講等比數(shù)列及其前n項(xiàng)和1.〖2021陜西百校聯(lián)考〗已知等比數(shù)列{an}的公比為q,前4項(xiàng)的和為a1+14,且a2,a3+1,a4成等差數(shù)列,則q的值()A.12或2 B.1或122.〖2021安徽省四校聯(lián)考〗已知正項(xiàng)等比數(shù)列{an}的前n項(xiàng)和為Sn,若a4=18,S3-a1=34,則S4=(A.116 B.18 C.313.〖2020合肥三檢〗〖數(shù)學(xué)文化題〗公元前1650年左右的埃及《萊因德紙草書(shū)》上載有如下問(wèn)題:“十人分十斗玉米,從第二人開(kāi)始,各人所得依次比前人少八分之一,問(wèn)每人各得玉米多少斗?”在上述問(wèn)題中,第一人分得玉米()A.70×89810-1斗 B.10×81084.〖2020南昌市測(cè)試〗公比不為1的等比數(shù)列{an}中,若a1a5=aman,則mn不可能為()A.5 B.6 C.8 D.95.〖2020成都市高三摸底測(cè)試〗已知等比數(shù)列{an}的各項(xiàng)均為正數(shù),若log3a1+log3a2+…+log3a12=12,則a6a7=()A.1 B.3 C.6 D.96.〖2021四省八校聯(lián)考〗已知等比數(shù)列{an}的公比為q,前n項(xiàng)和為Sn=m-qn,若a5=-8a2,則S5=.

7.〖2020大同市高三調(diào)研〗已知各項(xiàng)均為正數(shù)的等比數(shù)列{an}中,a1a2a3=5,a7a8a9=10,則a4a5a6=.

8.〖2020全國(guó)卷Ⅲ,17,12分〗設(shè)等比數(shù)列{an}滿足a1+a2=4,a3-a1=8.(1)求{an}的通項(xiàng)公式;(2)記Sn為數(shù)列{log3an}的前n項(xiàng)和.若Sm+Sm+1=Sm+3,求m.9.〖條件創(chuàng)新〗已知等比數(shù)列{an}的前n項(xiàng)和為Sn,且an+1+λ3=Sn,a3=12,則實(shí)數(shù)λA.-34 B.-14 C.10.〖設(shè)問(wèn)創(chuàng)新〗已知等比數(shù)列{an}的前5項(xiàng)積為32,1<a1<2,則a1+a32+a5A.〖3,72) B.(3,+∞) C.(3,72) D.11.〖2020成都市三診〗在等比數(shù)列{an}中,已知anan+1=9n,則該數(shù)列的公比是()A.-3 B.3 C.±3 D.912.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.1013.〖2021四省八校聯(lián)考〗設(shè)無(wú)窮數(shù)列{an}的前n項(xiàng)和為Sn,有三個(gè)條件:①am+n=am·an,②Sn=an+1+1,a1≠0,③Sn=2an+1p(p是與n無(wú)關(guān)的參數(shù)),則從中選出兩個(gè)條件,能使數(shù)列{an}為唯一確定的等比數(shù)列的條件是14.〖2020安徽省示范高中名校聯(lián)考〗設(shè)Sn是各項(xiàng)均為正數(shù)的等比數(shù)列{an}的前n項(xiàng)和,a1=3,若-a4,a3,a5成等差數(shù)列,則Sn與an的關(guān)系式為.

15.已知公比q>1的等比數(shù)列{an}滿足a52=a10,2(an+an+2)=5an+1.若bn=(n-λ)an(n∈N*),且數(shù)列{bn}是遞增數(shù)列,則實(shí)數(shù)λ的取值范圍是16.〖2020海南,18,12分〗已知公比大于1的等比數(shù)列{an}滿足a2+a4=20,a3=8.(1)求{an}的通項(xiàng)公式;(2)求a1a2-a2a3+…+(-1)n-1anan+1.答案第六章數(shù)列第三講等比數(shù)列及其前n項(xiàng)和1.A由題意知a1+a2+a3+a4=a1+14,且a2+a4=2(a3+1),兩式聯(lián)立,可解得a3=4,所以a2+a4=10,a2a4=a32=16,解得a2=2,a4=8或a2.D解法一設(shè)等比數(shù)列{an}的公比為q(q>0且q≠1),∵a4=18,S3-a1=34,∴a1(1-q3)1-q-解法二設(shè)等比數(shù)列{an}的公比為q(q>0且q≠1),∵S3-a1=a2+a3=a4q2+a4q=34,a4=18,∴q=12,∴a1=a4q3=1,S43.C設(shè)第i個(gè)人分到的玉米斗數(shù)為ai(i=1,2,…,9,10),則{an}是公比為78的等比數(shù)列.由題意知a1〖1-(78)4.B由等比數(shù)列的性質(zhì)可知,m+n=6,m∈N*,n∈N*,當(dāng)m=n=3時(shí),mn=9;當(dāng)m=4,n=2時(shí),mn=8;當(dāng)m=5,n=1時(shí),mn=5.故選B.5.D因?yàn)榈缺葦?shù)列{an}的各項(xiàng)均為正數(shù),所以log3a1+log3a2+…+log3a12=log3(a1·a2·…·a12)=log3(a6a7)6=12,所以(a6a7)6=312=96,所以a6a7=9,故選D.6.33由a5=-8a2得a1q4=-8a1q(a1≠0),解得q=-2,則Sn=a1(1-qn)1-q=a13-a137.52各項(xiàng)均為正數(shù)的等比數(shù)列{an}中,a1a2a3=a23=5,a7a8a9=a83=10,則a4a5a6=8.(1)設(shè){an}的公比為q,則an=a1qn-1.由已知得a1+a1q=4,所以{an}的通項(xiàng)公式為an=3n-1.(2)由(1)知log3an=n-1.故Sn=n(由Sm+Sm+1=Sm+3得m(m-1)+(m+1)m=(m+3)(m+2),即m2-5m-6=0.解得m=-1(舍去)或m=6.9.A由條件得an+1+λ=3Sn,當(dāng)n≥2時(shí),an+λ=3Sn-1,兩式相減,得an+1=4an,又a3=12,所以a2=3,a1=34,將n=1代入an+1+λ=3Sn,得a2+λ=3a1,得λ=-3410.C因?yàn)榈缺葦?shù)列{an}的前5項(xiàng)積為32,所以a35=32,解得a3=2,則a5=a32a1=4a1,a1+a32+a54=a1+1+111.B設(shè){an}的公比為q,根據(jù)題意知a1a2=9,a2a3=92,所以9=a2a3a1a2=a3a1=q2?q=±3.又a112.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)n-1.∴Tn=(-6)n×(12)0+1+2+…+(n-1)=(-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(2k-1),∴T2k+2T2k=36k+1×(12)(k+1)(2k13.①③在①中,令m=n=1,得a2=a12;在②中,Sn=an+1+1,當(dāng)n≥2時(shí),Sn-1=an+1,兩式相減,得an=an+1-an,即an+1=2an;在③中,Sn=2an+1p,Sn+1=2an+1+1p,兩式相減,得an+1=2an+1-2an,即an+1=2an.