2025屆高考數(shù)學(xué)一輪復(fù)習(xí)第5章數(shù)列第3講等比數(shù)列及其前n項(xiàng)和作業(yè)試題1含解析新人教版

PAGE第五章數(shù)列第三講等比數(shù)列及其前n項(xiàng)和練好題﹒考點(diǎn)自測1.[北京高考,5分]設(shè){an}是公比為q的等比數(shù)列,則“q>1”是“{an}為遞增數(shù)列”的 ()A.充分而不必要條件 B.必要而不充分條件C.充分必要條件 D.既不充分也不必要條件2.[2024全國卷Ⅲ,5分]已知各項(xiàng)均為正數(shù)的等比數(shù)列{an}的前4項(xiàng)和為15,且a5=3a3+4a1,則a3= ()A.16 B.8 C.4 D.23.[易錯(cuò)題]記等比數(shù)列{an}的前n項(xiàng)和為Sn,若a1=2,S3=6,則S4= ()A.10或8 B.-10C.-10或8 D.-10或-84.[2024全國卷Ⅱ,5分]數(shù)列{an}中,a1=2,am+n=aman.若ak+1+ak+2+…+ak+10=215-25,則k= ()A.2 B.3 C.4 D.55.[2024全國卷Ⅰ,5分]設(shè){an}是等比數(shù)列,且a1+a2+a3=1,a2+a3+a4=2,則a6+a7+a8= ()A.12 B.24 C.30 D.326.[2017全國卷Ⅱ,5分]我國古代數(shù)學(xué)名著《算法統(tǒng)宗》中有如下問題:“遠(yuǎn)望巍巍塔七層,紅光點(diǎn)點(diǎn)倍加增,共燈三百八十一,請問尖頭幾盞燈?”意思是:一座7層塔共掛了381盞燈,且相鄰兩層中的下一層燈數(shù)是上一層燈數(shù)的2倍,則塔的頂層共有燈 ()A.1盞 B.3盞 C.5盞 D.9盞7.[多選題]下列結(jié)論中,錯(cuò)誤的為 ()A.滿意an+1=qan(n∈N*,q為常數(shù))的數(shù)列{an}為等比數(shù)列B.a,b,c三個(gè)數(shù)成等比數(shù)列的必要條件是b2=acC.假如數(shù)列{an}為等比數(shù)列,bn=a2n-1+a2n,則數(shù)列{bn}也是等比數(shù)列D.假如數(shù)列{an}為等比數(shù)列,則數(shù)列{lnan}是等差數(shù)列8.[2016全國卷Ⅰ,5分]設(shè)等比數(shù)列{an}滿意a1+a3=10,a2+a4=5,則a1a2…an的最大值為.

拓展變式1.[2016全國卷Ⅲ,12分]已知數(shù)列{an}的前n項(xiàng)和Sn=1+λan,其中λ≠0.(1)證明{an}是等比數(shù)列,并求其通項(xiàng)公式;(2)若S5=QUOTE,求λ.2.(1)[2024全國卷Ⅱ,5分]記Sn為等比數(shù)列{an}的前n項(xiàng)和.若a5-a3=12,a6-a4=24,則QUOTE= ()A.2n-1B.2-21-nC.2-2n-1D.21-n-1(2)已知數(shù)列{an}是等差數(shù)列,若a2,a4+3,a6+6構(gòu)成公比為q的等比數(shù)列,則q= ()A.1 B.2 C.3 D.43.(1)[2024大同市調(diào)研測試]已知各項(xiàng)均為正數(shù)的等比數(shù)列{an}中,a1a2a3=5,a4a5a6=5QUOTE,則a7a8a9= ()A.25 B.20 C.10QUOTE D.10(2)已知正項(xiàng)等比數(shù)列{an}的前n項(xiàng)和為Sn,且S8-2S4=5,則a9+a10+a11+a12的最小值為 ()A.25 B.20 C.15 D.10(3)記等比數(shù)列{an}的前n項(xiàng)積為Tn(n∈N*),已知am-1am+1-2am=0,且T2m-1=128,則m的值為 ()A.4 B.7 C.10 D.124.數(shù)列{an},{bn}滿意a1=-1,b1=2,且QUOTE(n∈N*),則b2015+b2016=.

答案第三講等比數(shù)列及其前n項(xiàng)和1.D等比數(shù)列-1,-2,-4,…,滿意公比q=2>1,但{an}不是遞增數(shù)列,即充分性不成立.an=-1×(QUOTE)n-1為遞增數(shù)列,但q=QUOTE<1,即必要性不成立.故“q>1”是“{an}為遞增數(shù)列”的既不充分也不必要條件,故選D.2.C設(shè)等比數(shù)列{an}的公比為q,由a5=3a3+4a1得q4=3q2+4,得q2=4,因?yàn)閿?shù)列{an}的各項(xiàng)均為正數(shù),所以q=2.又a1+a2+a3+a4=a1(1+q+q2+q3)=a1(1+2+4+8)=15,所以a1=1,所以a3=a1q2=4,故選C.3.C設(shè)等比數(shù)列的公比為q,因?yàn)閍1=2,S3=6,所以S3=2+2q+2q2=6,則q2+q-2=0,所以q=1或q=-2.當(dāng)q=1時(shí),S4=S3+2=8;當(dāng)q=-2時(shí),S4=S3+a1q3=6+2×(-2)3=-10.故選C.4.C令m=1,則由am+n=aman,得an+1=a1an,即QUOTE=a1=2,所以數(shù)列{an}是首項(xiàng)為2、公比為2的等比數(shù)列,所以an=2n,所以ak+1+ak+2+…+ak+10=ak(a1+a2+…+a10)=2k×QUOTE=2k+1×(210-1)=215-25=25×(210-1),解得k=4,故選C.5.D解法一設(shè)等比數(shù)列{an}的公比為q,所以QUOTE=QUOTE=q=2,由a1+a2+a3=a1(1+q+q2)=a1(1+2+22)=1,解得a1=QUOTE,所以a6+a7+a8=a1(q5+q6+q7)=QUOTE×(25+26+27)=QUOTE×25×(1+2+22)=32,故選D.解法二令bn=an+an+1+an+2(n∈N*),則bn+1=an+1+an+2+an+3.設(shè)數(shù)列{an}的公比為q,則QUOTE=QUOTE=QUOTE=q,所以數(shù)列{bn}為等比數(shù)列,由題意知b1=1,b2=2,所以等比數(shù)列{bn}的公比q=2,所以bn=2n-1,所以b6=a6+a7+a8=25=32,故選D.6.B由題意知,每層塔所掛的燈數(shù)從上到下構(gòu)成等比數(shù)列,記為{an},則前7項(xiàng)的和S7=381,公比q=2,依題意,得QUOTE=381,解得a1=3,故選B.7.ACD對于A,當(dāng)n屬于正整數(shù),q為常數(shù)且不等于0時(shí),數(shù)列{an}為等比數(shù)列,故A錯(cuò)誤;對于B,由等比中項(xiàng)的概念可知,a,b,c三個(gè)數(shù)成等比數(shù)列的必要條件是b2=ac,故B正確;對于C,當(dāng)?shù)缺葦?shù)列{an}的公比q=-1時(shí),bn=0,此時(shí){bn}不是等比數(shù)列,故C錯(cuò)誤;對于D,當(dāng)an為正數(shù)時(shí),數(shù)列{lnan}是等差數(shù)列,故D錯(cuò)誤.8.64解法一設(shè)等比數(shù)列{an}的公比為q,由QUOTE得,QUOTE解得QUOTE所以a1a2…an=QUOTEq1+2+…+(n-1)=8n×(QUOTE=QUOTE.記t=-QUOTE+QUOTE(n∈N*),則t=-QUOTE(n2-7n)=-QUOTE(n-QUOTE)2+QUOTE,易知當(dāng)n=3或4時(shí),a1a2…an取得最大值26=64.解法二設(shè)等比數(shù)列{an}的公比為q,由題可知a1+a1q2=10,a1q+a1q3=5,則q=QUOTE,a1=8,所以an=8·(QUOTE)n-1=24-n.當(dāng)n=4時(shí),a4=1,所以a1>a2>a3>a4=1>a5>a6>….所以a1a2…an取最大值時(shí)n=3或4.所以a1a2…an的最大值為64.1.(1)由題意得a1=S1=1+λa1,故λ≠1,a1=QUOTE,a1≠0.由Sn=1+λan,Sn+1=1+λan+1得an+1=λan+1-λan,即an+1(λ-1)=λan.又a1≠0,λ≠0且λ≠1,所以QUOTE=QUOTE.所以{an}是首項(xiàng)為QUOTE,公比為QUOTE的等比數(shù)列,故an=QUOTE(QUOTE)n-1.(2)由(1)得Sn=1-(QUOTE)n.由S5=QUOTE得1-(QUOTE)5=QUOTE,即(QUOTE)5=QUOTE,解得λ=-1.2.(1)B解法一設(shè)等比數(shù)列{an}的公比為q,則由QUOTE解得QUOTE所以Sn=QUOTE=2n-1,an=a1qn-1=2n-1,所以QUOTE=QUOTE=2-21-n,故選B.解法二設(shè)等比數(shù)列{an}的公比為q,因?yàn)镼UOTE=QUOTE=QUOTE=QUOTE=2,所以q=2,所以QUOTE=QUOTE=QUOTE=2-21-n,故選B.(2)A設(shè)等差數(shù)列{an}的公差為d,由a2,a4+3,a6+6構(gòu)成公比為q的等比數(shù)列,得(a4+3)2=a2(a6+6),即(a1+3d+3)2=(a1+d)·(a1+5d+6),化簡得(2d+3)2=0,解得d=-QUOTE.所以q=QUOTE=QUOTE=QUOTE=1.故選A.3.(1)D解法一因?yàn)閿?shù)列{an}為正項(xiàng)等比數(shù)列,所以a1a2a3=QUOTE=5,a4a5a6=QUOTE=5QUOTE,a7a8a9=QUOTE,又a2a8=QUOTE,所以QUOTE=QUOTE=QUOTE=50,得a7a8a9=QUOTE=QUOTE=10,故選D.解法二因?yàn)閿?shù)列{an}為正項(xiàng)等比數(shù)列,所以a1a2a3,a4a5a6,a7a8a9也構(gòu)成等比數(shù)列,所以QUOTE=(a1a2a3)·(a7a8a9)=50,又a1a2a3=5,所以a7a8a9=10,故選D.(2)B在正項(xiàng)等比數(shù)列{an}中,Sn>0.因?yàn)镾8-2S4=5,所以S8-S4=5+S4.易知S4,S8-S4,S12-S8成等比數(shù)列,所以QUOTE=S4·(S12-S8),所以S12-S8=QUOTE=QUOTE+S4+10≥2QUOTE+10=20(當(dāng)且僅當(dāng)S4=5時(shí)取等號).因?yàn)镾12-S8=a9+a10+a11+a12,所以a9+a10+a11+a12的最小值為20.故選B.(3)A因?yàn)閧an}是等比數(shù)列,所以am-1am+1=QUOTE.又am-1am+1-2am=0,則QUOTE-2am=0,所以am=2(am=0舍去).由等比數(shù)列