（广西课标版）2020版高考数学二轮复习专题能力训练11等差数列与等比数列文.docx

专题能力训练11等差数列与等比数列一、能力突破训练1.已知数列an为等比数列,且a8a9a10=-a132=-1 000,则a10a12=()A.100B.-100C.10010D.-100102.在等差数列an中,a1+a2+a3=3,a18+a19+a20=87,则此数列前20项的和等于()A.290B.300C.580D.6003.设an是等比数列,Sn是an的前n项和.若对任意正整数n,有an+2an+1+an+2=0,a1=2,则S101的值为()A.2B.200C.-2D.04.已知an是等差数列,公差d不为零,前n项和是Sn.若a3,a4,a8成等比数列,则()A.a1d0,dS40B.a1d0,dS40,dS40D.a1d05.在等比数列an中,满足a1+a2+a3+a4+a5=3,a12+a22+a32+a42+a52=15,则a1-a2+a3-a4+a5的值是()A.3B.5C.-5D.56.在数列an中,a1=2,an+1=2an,Sn为an的前n项和.若Sn=126,则n=.7.(2019四川内江等六市二诊,14)中国古代数学专著九章算术中有这样一题:今有男子善走,日增等里,九日走1 260里,第一日、第四日、第七日所走之和为390里,则该男子第三日走的里数为.8.设x,y,z是实数,若9x,12y,15z成等比数列,且1x,1y,1z成等差数列,则xz+zx=.9.(2018全国,文17)在等比数列an中,a1=1,a5=4a3.(1)求an的通项公式;(2)记Sn为an的前n项和,若Sm=63,求m.10.(2019全国,文18)记Sn为等差数列an的前n项和.已知S9=-a5.(1)若a3=4,求an的通项公式;(2)若a10,求使得Snan的n的取值范围.11.(2019山东潍坊四市联考,17)已知数列an,bn满足:an+1+1=2an+n,bn-an=n,b1=2.(1)证明数列bn是等比数列,并求数列bn的通项公式;(2)求数列an的前n项和Sn.二、思维提升训练12.已知数列an,bn满足a1=b1=1,an+1-an=bn+1bn=2,nN*,则数列ban的前10项的和为()A.43(49-1)B.43(410-1)C.13(49-1)D.13(410-1)13.若数列an为等比数列,且a1=1,q=2,则Tn=1a1a2+1a2a3+1anan+1等于()A.1-14nB.231-14nC.1-12nD.231-12n14.如图,点列An,Bn分别在某锐角的两边上,且|AnAn+1|=|An+1An+2|,AnAn+2,nN*,|BnBn+1|=|Bn+1Bn+2|,BnBn+2,nN*.(PQ表示点P与Q不重合)若dn=|AnBn|,Sn为AnBnBn+1的面积,则()A.Sn是等差数列B.Sn2是等差数列C.dn是等差数列D.dn2是等差数列15.(2019河北武邑中学调研,15)若两个等差数列an和bn的前n项和分别是Sn,Tn,SnTn=5nn+5,则a10b9+b12+a11b8+b13=.16.(2019江苏常州高三期末,19)在数列an中,a1=1,且an+1+3an+4=0,nN*.(1)求证:an+1是等比数列,并求数列an的通项公式;(2)数列an中是否存在不同的三项按照一定顺序重新排列后,构成等差数列?若存在,求满足条件的项;若不存在,请说明理由.17.若数列an是公差为正数的等差数列,且对任意nN*有anSn=2n3-n2.(1)求数列an的通项公式;(2)是否存在数列bn,使得数列anbn的前n项和为An=5+(2n-3)2n-1(nN*)?若存在,求出数列bn的通项公式及其前n项和Tn;若不存在,请说明理由.专题能力训练11等差数列与等比数列一、能力突破训练1.C解析an为等比数列,a8a9a10=-a132=a93=-1000,a9=-10,a132=1000.又a10a12=a102q20,a10a12=|a9a13|=10010.2.B解析由a1+a2+a3=3,a18+a19+a20=87,得a1+a20=30,故S20=20(a1+a20)2=300.3.A解析设公比为q,an+2an+1+an+2=0,a1+2a2+a3=0,a1+2a1q+a1q2=0,q2+2q+1=0,q=-1.又a1=2,S101=a1(1-q101)1-q=21-(-1)1011+1=2.4.B解析设an的首项为a1,公差为d,则a3=a1+2d,a4=a1+3d,a8=a1+7d.a3,a4,a8成等比数列,(a1+3d)2=(a1+2d)(a1+7d),即3a1d+5d2=0.d0,a1d=-53d20,且a1=-53d.dS4=4d(a1+a4)2=2d(2a1+3d)=-23d20知d0,故Snan等价于n2-11n+100,解得1n10.所以n的取值范围是n|1n10,nN.11.解(1)因为bn-an=n,所以bn=an+n.因为an+1=2an+n-1,所以an+1+(n+1)=2(an+n),即bn+1=2bn.又b1=2,所以bn是首项为2,公比为2的等比数列,bn=22n-1=2n.(2)由(1)可得an=bn-n=2n-n,所以Sn=(21+22+23+2n)-(1+2+3+n)=2(1-2n)1-2-n(1+n)2=2n+1-2-n2+n2.二、思维提升训练12.D解析由a1=1,an+1-an=2,得an=2n-1.由bn+1bn=2,b1=1得bn=2n-1.则ban=2an-1=22(n-1)=4n-1,故数列ban的前10项和为1-4101-4=13(410-1).13.B解析因为an=12n-1=2n-1,所以anan+1=2n-12n=22n-1=24n-1,所以1anan+1=1214n-1.所以1anan+1是等比数列.故Tn=1a1a2+1a2a3+1anan+1=1211-14n1-14=231-14n.14.A解析如图,延长AnA1,BnB1交于P,过An作对边BnBn+1的垂线,其长度记为h1,过An+1作对边Bn+1Bn+2的垂线,其长度记为h2,则Sn=12|BnBn+1|h1,Sn+1=12|Bn+1Bn+2|h2.Sn+1-Sn=12|Bn+1Bn+2|h2-12|BnBn+1|h1.|BnBn+1|=|Bn+1Bn+2|,Sn+1-Sn=12|BnBn+1|(h2-h1).设此锐角为,则h2=|PAn+1|sin,h1=|PAn|sin,h2-h1=sin(|PAn+1|-|PAn|)=|AnAn+1|sin.Sn+1-Sn=12|BnBn+1|AnAn+1|sin.|BnBn+1|,|AnAn+1|,sin均为定值,Sn+1-Sn为定值.Sn是等差数列.故选A.15.4解析由等差数列的性质可得a10b9+b12+a11b8+b13=a10b1+b20+a11b1+b20=a1+a20b1+b20=20(a1+a20)220(b1+b20)2=S20T20=52020+5=4.16.解(1)因为an+1+3an+4=0,所以an+1+1an+1=-3an-3an+1=-3.因为a1+1=20,所以数列an+1是以2为首项,以-3为公比的等比数列,所以an+1=2(-3)n-1,即an=2(-3)n-1-1.(2)假设存在三项ar,as,at(rs0,an=dn+(a1-d),Sn=12dn2+a1-12dn.对任意nN*,恒有anSn=2n3-n2,则dn+(a1-d)12dn2+a1-12dn=2n3-n2,即dn+(a1-d)12dn+a1-12d=2n2-n.12d2=2,12d(a1-d)+da1-12d=-1,(a1-d)a1-12d=0.d0,a1=1,d=2,an=2n-1.(2)数列anbn的前n项和为An=5+(2n-3)2n-1(nN*),当n=1时,a1b1=