高考數(shù)學(xué)二輪總復(fù)習(xí)專題能力訓(xùn)練11等差數(shù)列與等比數(shù)列文

專題能力訓(xùn)練11等差數(shù)列與等比數(shù)列能力突破訓(xùn)練1.已知數(shù)列{an}為等比數(shù)列,且a8a9a10=a132=1000,則a10a12=(A.100 B.100 C.10010 D.100102.已知等差數(shù)列{an}的前n項(xiàng)和為Sn,當(dāng)首項(xiàng)a1和公差d變化時(shí),a3+a8+a10是一個(gè)定值,則下列選項(xiàng)中為定值的是()A.S7 B.S8 C.S13 D.S153.(2022廣西南寧二中高三檢測(cè))已知等比數(shù)列{an}滿足an>0,且a5·a2n5=e2n(n≥3),則當(dāng)n≥1時(shí),lna2+lna4+…+lna2n=()A.n(n+1) B.n(2n1)C.n2 D.(n1)24.已知{an}是等差數(shù)列,公差d不為零,前n項(xiàng)和是Sn.若a3,a4,a8成等比數(shù)列,則()A.a1d>0,dS4>0 B.a1d<0,dS4<0C.a1d>0,dS4<0 D.a1d<0,dS4>05.(2022廣西賀州模擬)設(shè)數(shù)列{an}的前n項(xiàng)和為Sn,已知2an+1+an=0,S5=112,則數(shù)列{an}的前n項(xiàng)之積Tn的最大值為(A.16 B.32 C.64 D.1286.已知Sn為等差數(shù)列{an}的前n項(xiàng)和,且S3=15,a3+a4+a5=27,則S10=.

7.(2022廣西河池高中模擬)若a,b,2這三個(gè)數(shù)經(jīng)適當(dāng)排序后可成等差數(shù)列,也可適當(dāng)排序后成等比數(shù)列,請(qǐng)寫出滿足題意的一組a,b的值:.

8.已知Sn是數(shù)列{an}的前n項(xiàng)和,a1=2,數(shù)列{log2Sn}是公差為2的等差數(shù)列,則S5=.

9.設(shè)Sn為數(shù)列{an}的前n項(xiàng)和.已知2Snn+n=2an(1)證明:{an}是等差數(shù)列;(2)若a4,a7,a9成等比數(shù)列,求Sn的最小值.10.已知在數(shù)列{an}中,a1=1,2an+1=6an+2n1.(1)求證:數(shù)列an(2)求數(shù)列{an}的前n項(xiàng)和Sn.11.已知數(shù)列{an},{bn}滿足:an+1+1=2an+n,bnan=n,b1=2.(1)證明數(shù)列{bn}是等比數(shù)列,并求數(shù)列{bn}的通項(xiàng)公式;(2)求數(shù)列{an}的前n項(xiàng)和Sn.思維提升訓(xùn)練12.已知數(shù)列{an},{bn}滿足a1=b1=1,an+1an=bn+1bn=2,n∈N*,則數(shù)列{A.43(491) B.43(4C.13(491) D.13(413.已知等差數(shù)列{an}的公差為2,前n項(xiàng)和為Sn,且S1,S2,S4成等比數(shù)列.令bn=1anan+1,則數(shù)列{bn}的前50項(xiàng)和TA.5051 B.49C.100101 D.14.將數(shù)列{2n1}與{3n2}的公共項(xiàng)從小到大排列得到數(shù)列{an},則{an}的前n項(xiàng)和為.

15.若兩個(gè)等差數(shù)列{an}和{bn}的前n項(xiàng)和分別是Sn,Tn,SnTn=5n16.已知數(shù)列{bn}是各項(xiàng)均為正數(shù)的等比數(shù)列,其前n項(xiàng)和為Tn,b1=2,T4=5T2.等差數(shù)列{an}的前n項(xiàng)和為Sn,且a1+a3=b3,a1+a9=4.(1)求{an}和{bn}的通項(xiàng)公式.(2)是否存在大于2的正整數(shù)m,使得4S1,S3,Sm成等比數(shù)列?若存在,求出m的值;若不存在,請(qǐng)說明理由.17.已知等差數(shù)列{an}滿足a1=1,公差d>0,等比數(shù)列{bn}滿足b1=a1,b2=a2,b3=a5.(1)求數(shù)列{an},{bn}的通項(xiàng)公式;(2)若數(shù)列{cn}滿足c1b1+c2b2+c3b3+…+答案：能力突破訓(xùn)練1.C解析:∵{an}為等比數(shù)列,∴a8a9a10=a132∴a9=10,a132=又a10a12=a102q2∴a10a12=|a9a13|=10010.2.C解析:由題意,可知a3+a8+a10=3a1+18d=3(a1+6d)=3a7是一個(gè)定值,所以a7是一個(gè)定值,所以S13=13(a1+a133.A解析:由a5·a2n5=e2n(n≥3)得an2=e2n(又an>0,所以an=en(n≥3).又?jǐn)?shù)列{an}為等比數(shù)列,所以an=en,所以當(dāng)n≥1時(shí),lna2+lna4+…+lna2n=2+4+…+2n=n(n+1).4.B解析:∵a3=a1+2d,a4=a1+3d,a8=a1+7d,a3,a4,a8成等比數(shù)列,∴(a1+3d)2=(a1+2d)(a1+7d),即3a1d+5d2=0.∵d≠0,∴a1d=53d2<0,且a1=53∴dS4=4d(a1+a4)2=2d(2a1+故選B.5.C解析:由2an+1+an=0,S5=112可知,an≠0,所以an+1所以數(shù)列{an}是以12為公比的等比數(shù)列由S5=112可知,a11--1所以數(shù)列{an}為8,4,2,1,12,1所以{an}的前n項(xiàng)之積Tn的最大值為T4=8×(4)×2×(1)=64.6.120解析:設(shè)等差數(shù)列{an}的公差為d,根據(jù)題意得3解得a1=3,d=2,所以S10=10a1+10×927.a=1,b=4(答案不唯一)解析:a,b,2這三個(gè)數(shù)經(jīng)適當(dāng)排序后可成等差數(shù)列,可排為2,a,b,則有2+b=2a.a,b,2這三個(gè)數(shù)經(jīng)適當(dāng)排序后可成等比數(shù)列,可排為a,2,b,則有ab=(2)2.所以a=1,b=4或a=2,b=2.其他情況解法同上.8.512解析:∵數(shù)列{log2Sn}是公差為2的等差數(shù)列,∴l(xiāng)og2Snlog2Sn1=2,即SnSn-1=4(n≥2),又S∴數(shù)列{Sn}是公比為4的等比數(shù)列,則Sn=2×4n1,∴S5=2×44=512.9.(1)證明:由2Snn+n=2an+1,變形為2Sn=2nan+nn2,記為①式,又當(dāng)n≥2時(shí),有2Sn1=2(n1)an1+n1(n1)2,記為②式,①②并整理可得(2n2)an(2n2)an1=2n2,n≥2,n∈所以anan1=1,n≥2,n∈N*,所以{an}是等差數(shù)列.(2)解:由題意可知a72=a4a9,即(a1+6)2=(a1+3)·(a1+8),解得a1=12,所以an=12+(n1)×1=n13,當(dāng)n≤12時(shí),an<0,當(dāng)n=13時(shí),an=0,當(dāng)n≥14時(shí),an>0.故Sn的最小值為S12=S13=10.(1)證明:∵2an+1=6an+2n1,∴an+1=3an+n12∴an+1+n+12=3又a1+12=1+12=32,∴(2)解:由(1)得an+n2=32×3n∴an=3n∴Sn=a1+a2+a3+…+an=12(31+32+33+…+3n)12(1+2+3+…=1211.解:(1)因?yàn)閎nan=n,所以bn=an+n.因?yàn)閍n+1=2an+n1,所以an+1+(n+1)=2(an+n),即bn+1=2bn.又b1=2,所以{bn}是首項(xiàng)為2,公比為2的等比數(shù)列,bn=2×2n1=2n.(2)由(1)可得an=bnn=2nn,所以Sn=(21+22+23+…+2n)(1+2+3+…+n)=2(1-2n)1思維提升訓(xùn)練12.D解析:由a1=1,an+1an=2,得an=2n1.由bn+1bn=2,b1=1得bn=則ban=2an-1=2故數(shù)列{ban}的前10項(xiàng)和為1-413.D解析:因?yàn)镾1=a1,S2=2a1+2×12×2=2a1+2,S4=4a1+4×32×2由題意得(2a1+2)2=a1(4a1+12),解得a1=1,所以an=2n1,則bn=1(2n-1)(2n14.3n22n解析:數(shù)列{2n1}的項(xiàng)均為奇數(shù),數(shù)列{3n2}的所有奇數(shù)項(xiàng)均為奇數(shù),所有偶數(shù)項(xiàng)均為偶數(shù),并且顯然{3n2}中的所有奇數(shù)均能在{2n1}中找到,所以{2n1}與{3n2}的所有公共項(xiàng)就是{3n2}的所有奇數(shù)項(xiàng),這些項(xiàng)從小到大排列得到的新數(shù)列{an}為以1為首項(xiàng),以6為公差的等差數(shù)列.所以{an}的前n項(xiàng)和為Sn=n×1+n(n-1)2×6=15.4解析:由等差數(shù)列的性質(zhì)可得a10b916.解:(1)設(shè){bn}的公比為q(q>0).當(dāng)q=1時(shí),顯然不滿足T4=5T2,故q≠1.所以由T4=5T2,得b1(1-q4又q>0,所以q=2,所以bn=2n.設(shè){an}的公差為d.由a解得a1=6,d=-2.(2)因?yàn)镾n=n(6-2n+8)2=n2+7n,所以S1=6,S3=12,若4S1,S3,Sm成等比數(shù)列,則S32=4S1S即122=24(m2+7m),化簡(jiǎn)得m27m+6=0,解得m=6或m=1(舍去).故存在m=6,使得4S1,S3,Sm成等比數(shù)列.17.解:(1)由題意知,a1=1,公差d>0,有1,1+d,1+4d成等比數(shù)列,所以(1+d)2=1×(1+4