PAGE2025年高考數(shù)學(xué)一輪復(fù)習(xí)課時(shí)作業(yè)-等比數(shù)列【原卷版】(時(shí)間:45分鐘分值:85分)【基礎(chǔ)落實(shí)練】1.(5分)在等比數(shù)列{an}中,a1a3=a4=4,則a6=()A.6 B.-8或82.(5分)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.843.(5分)已知等比數(shù)列{an}的前n項(xiàng)和為Sn=a·2n-1+16,則a的值為(A.-13 B.13 C.-12 4.(5分)已知各項(xiàng)均為正數(shù)的等比數(shù)列{an}中,a4與a14的等比中項(xiàng)為22,則log2a7+log2a11的值為()A.1 B.2 C.3 D.4【加練備選】設(shè)等比數(shù)列{an}的公比為q>0,且q≠1,Sn為數(shù)列{an}的前n項(xiàng)和,記Tn=anSnA.T3≤T6 B.T3<T6C.T3≥T6 D.T3>T65.(5分)已知數(shù)列{an}為等比數(shù)列,且a2a10=4a6,Sn為等差數(shù)列{bn}的前n項(xiàng)和,且S6=S10,a6=b7,則b9=()A.43 B.-43 C.-83 6.(5分)(多選題)設(shè)等比數(shù)列{an}的公比為q,則下列說法正確的是()A.數(shù)列{anan+1}是公比為q2的等比數(shù)列B.數(shù)列{an+an+1}是公比為q的等比數(shù)列C.數(shù)列{an-an+1}是公比為q的等比數(shù)列D.數(shù)列1an是公比為7.(5分)若等差數(shù)列{an}和等比數(shù)列{bn}滿足a1=b1=-1,a4=b4=8,則a2b28.(5分)已知數(shù)列{an}是等比數(shù)列,a2=2,a5=14則a1a2a3+a2a3a4+…+anan+1an+2=.

9.(5分)已知{an}是各項(xiàng)均為正數(shù)的等比數(shù)列,Sn為其前n項(xiàng)和.若a1=6,a2+2a3=6,則公比q=,S4=.

10.(10分)在數(shù)列{an}中,已知a1=1,an+1=2an+2n-3.(1)若bn=an+2n-1,證明:數(shù)列{bn}是等比數(shù)列.(2)求數(shù)列{an}的前n項(xiàng)和Sn.11.(10分)(2022·新高考Ⅱ卷)已知{an}是等差數(shù)列,{bn}是公比為2的等比數(shù)列,且a2-b2=a3-b3=b4-a4.(1)證明:a1=b1;(2)求集合{k|bk=am+a1,1≤m≤500}中元素的個(gè)數(shù).【能力提升練】12.(5分)(多選題)設(shè)等比數(shù)列{an}的公比為q,其前n項(xiàng)和為Sn,前n項(xiàng)積為Tn,并且滿足條件a1>1,a9a10>1,a9-1A.0<q<1 B.a10a11>1C.Sn的最大值為S10 D.Tn的最大值為T913.(5分)等比數(shù)列{an}的首項(xiàng)a1=-1,前n項(xiàng)和為Sn,若S10S5=242243,則公比14.(10分)已知公比不為1的等比數(shù)列{an}滿足a1+a3=5,且a1,a3,a2構(gòu)成等差數(shù)列.(1)求{an}的通項(xiàng)公式;(2)記Sn為{an}的前n項(xiàng)和,求使Sk>238成立的最大正整數(shù)k的值2025年高考數(shù)學(xué)一輪復(fù)習(xí)課時(shí)作業(yè)-等比數(shù)列【解析版】(時(shí)間:45分鐘分值:85分)【基礎(chǔ)落實(shí)練】1.(5分)在等比數(shù)列{an}中,a1a3=a4=4,則a6=()A.6 B.-8或8C.-8 D.8【解析】選D.因?yàn)閍1·a3=a22=4,所以a2當(dāng)a2=-2時(shí),a32=a2·a所以a2=2,所以q2=a4所以a6=a4·q2=4×2=8.2.(5分)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.84【解析】選B.設(shè)數(shù)列{an}的公比為q,則a1(1+q2+q4)=21,又a1=3,所以q4+q2-6=0,所以q2=2(q2=-3舍去),所以a3=6,a5=12,a7=24,所以a3+a5+a7=42.3.(5分)已知等比數(shù)列{an}的前n項(xiàng)和為Sn=a·2n-1+16,則a的值為(A.-13 B.13 C.-12 【解析】選A.當(dāng)n≥2時(shí),an=Sn-Sn-1=a·2n-1-a·2n-2=a·2n-2,當(dāng)n=1時(shí),a1=S1=a+16又因?yàn)閿?shù)列{an}是等比數(shù)列,所以a+16=a2,所以a=-4.(5分)已知各項(xiàng)均為正數(shù)的等比數(shù)列{an}中,a4與a14的等比中項(xiàng)為22,則log2a7+log2a11的值為()A.1 B.2 C.3 D.4【解析】選C.由題意得a4a14=(22)2=8,由等比數(shù)列的性質(zhì),得a4a14=a7a11=8,所以log2a7+log2a11=log2(a7a11)=log28=3.【加練備選】設(shè)等比數(shù)列{an}的公比為q>0,且q≠1,Sn為數(shù)列{an}的前n項(xiàng)和,記Tn=anSnA.T3≤T6 B.T3<T6C.T3≥T6 D.T3>T6【解析】選D.T6-T3=a6(1-q)a1(由于q>0且q≠1,所以1-q與1-q6同號,所以T6-T3<0,所以T6<T3.5.(5分)已知數(shù)列{an}為等比數(shù)列,且a2a10=4a6,Sn為等差數(shù)列{bn}的前n項(xiàng)和,且S6=S10,a6=b7,則b9=()A.43 B.-43 C.-83 【解析】選B.因?yàn)閿?shù)列{an}為等比數(shù)列,且a2a10=4a6,所以a62=4a6,解得a6設(shè)等差數(shù)列{bn}的公差為d,因?yàn)镾6=S10,所以b7+b8+b9+b10=0,則b7+b10=0.因?yàn)閍6=b7=4,所以b10=-4,所以3d=b10-b7=-4-4=-8,所以d=-83所以b9=b7+2d=4+2×(-83)=-46.(5分)(多選題)設(shè)等比數(shù)列{an}的公比為q,則下列說法正確的是()A.數(shù)列{anan+1}是公比為q2的等比數(shù)列B.數(shù)列{an+an+1}是公比為q的等比數(shù)列C.數(shù)列{an-an+1}是公比為q的等比數(shù)列D.數(shù)列1an是公比為【解析】選AD.對于A,由anan+1an-1an=q2(n對于B,當(dāng)q=-1時(shí),數(shù)列{an+an+1}的項(xiàng)中有0,不是等比數(shù)列;對于C,當(dāng)q=1時(shí),數(shù)列{an-an+1}的項(xiàng)中有0,不是等比數(shù)列;對于D,1an+11a所以數(shù)列1an是公比為17.(5分)若等差數(shù)列{an}和等比數(shù)列{bn}滿足a1=b1=-1,a4=b4=8,則a2b2【解析】設(shè)等差數(shù)列{an}的公差為d,等比數(shù)列{bn}的公比為q.由題意得-1+3d=-q3=8?d=3,q=-2?a2b2=答案:18.(5分)已知數(shù)列{an}是等比數(shù)列,a2=2,a5=14則a1a2a3+a2a3a4+…+anan+1an+2=.

【解析】設(shè)數(shù)列{an}的公比為q,則q3=a5a2=18,解得q=12,a1=a2q=4,a易知數(shù)列{anan+1an+2}是首項(xiàng)為a1a2a3=4×2×1=8,公比為q3=18所以a1a2a3+a2a3a4+…+anan+1an+2=8(1-18n)1-答案:647(1-2-3n9.(5分)已知{an}是各項(xiàng)均為正數(shù)的等比數(shù)列,Sn為其前n項(xiàng)和.若a1=6,a2+2a3=6,則公比q=,S4=.

【解析】由題意,數(shù)列{an}是各項(xiàng)均為正數(shù)的等比數(shù)列,由a1=6,a2+2a3=6,可得a1q+2a1q2=6q+12q2=6,即2q2+q-1=0,解得q=12或q=-1(舍去)由等比數(shù)列的前n項(xiàng)和公式,可得S4=6×[1-(1答案:1210.(10分)在數(shù)列{an}中,已知a1=1,an+1=2an+2n-3.(1)若bn=an+2n-1,證明:數(shù)列{bn}是等比數(shù)列.(2)求數(shù)列{an}的前n項(xiàng)和Sn.【解析】(1)因?yàn)閍n+1=2an+2n-3,bn=an+2n-1,所以bn+1bn=a又b1=a1+2-1=2,所以數(shù)列{bn}是以2為首項(xiàng),2為公比的等比數(shù)列.(2)由(1)可知,bn=2n,則an=2n-2n+1,Sn=21-1+22-3+…+2n-2n+1=21+22+…+2n-(1+3+…+2n-1)=2-2n+11-2-n(11.(10分)(2022·新高考Ⅱ卷)已知{an}是等差數(shù)列,{bn}是公比為2的等比數(shù)列,且a2-b2=a3-b3=b4-a4.(1)證明:a1=b1;(2)求集合{k|bk=am+a1,1≤m≤500}中元素的個(gè)數(shù).【解析】(1)設(shè)等差數(shù)列{an}的公差為d,由a2-b2=a3-b3得a1+d-2b1=a1+2d-4b1,即d=2b1,由a2-b2=b4-a4得a1+d-2b1=8b1-(a1+3d),即a1=5b1-2d,將d=2b1代入,得a1=5b1-2×2b1=b1,即a1=b1.【解析】(2)由(1)知an=a1+(n-1)d=a1+(n-1)×2b1=(2n-1)a1,bn=b1·2n-1,由bk=am+a1,得b1·2k-1=(2m-1)a1+a1,由a1=b1≠0得2k-1=2m,由題知1≤m≤500,所以2≤2m≤1000,所以k=2,3,4,…,10,共9個(gè)數(shù),即集合{k|bk=am+a1,1≤m≤500}={2,3,4,…,10}中元素的個(gè)數(shù)為9.【能力提升練】12.(5分)(多選題)設(shè)等比數(shù)列{an}的公比為q,其前n項(xiàng)和為Sn,前n項(xiàng)積為Tn,并且滿足條件a1>1,a9a10>1,a9-1A.0<q<1 B.a10a11>1C.Sn的最大值為S10 D.Tn的最大值為T9【解析】選AD.由題意得a9>1>a10>a11…,所以0<q<1,a10a11<1,Sn沒有最大值,T9最大.13.(5分)等比數(shù)列{an}的首項(xiàng)a1=-1,前n項(xiàng)和為Sn,若S10S5=242243,則公比【解析】由S10S5=242243知公比q≠1,S10-S由等比數(shù)列前n項(xiàng)和的性質(zhì)知S5,S10-S5,S15-S10成等比數(shù)列,且公比為q5,故q5=-1243,所以q=-1答案:-114.(10分)已知公比不為1的等比數(shù)列{an}滿足a1+a3=5,且a1,a3,a2構(gòu)成等差數(shù)列.(1)求{an}的通項(xiàng)公式;(2)記Sn為{an}的前n項(xiàng)和,求使