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专题能力训练11等差数列与等比数列一、能力突破训练1.已知等比数列an满足a1=,a3a5=4(a4-1),则a2=()A.2B.1C.D.2.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.对任意正整数n,有an+2an+1+an+2=0,又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.已知等比数列an为递增数列,且a52=a10,2(an+an+2)=5an+1,则数列的通项公式an=.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.已知等差数列an和等比数列bn满足a1=b1=1,a2+a4=10,b2b4=a5.(1)求an的通项公式;(2)求和:b1+b3+b5+b2n-1.11.设数列an满足a1+3a2+(2n-1)an=2n.(1)求an的通项公式;(2)求数列an2n+1的前n项和.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. (49-1)B. (410-1)C. (49-1)D. (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.已知等比数列an的首项为,公比为-,其前n项和为Sn,若ASn-1SnB对nN*恒成立,则B-A的最小值为.16.已知数列an的首项为1,Sn为数列an的前n项和,Sn+1=qSn+1,其中q0,nN*.(1)若a2,a3,a2+a3成等差数列,求数列an的通项公式;(2)设双曲线x2-y2an2=1的离心率为en,且e2=2,求e12+e22+en2.17.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 a3a5=4(a4-1),a42=4(a4-1),解得a4=2.又a4=a1q3,且a1=14,q=2,a2=a1q=12.2.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,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=A1=4,b1=4,当n2时,anbn=An-An-1=5+(2n-3)2n-1-5+(2n-5)2n-2=(2n-1)2n-2.bn=2n-2.假设存在数列bn满足题设,且数列bn的通项公式bn=4,n=1,2n-2,n2,T1=4,当n2时,Tn=4+1-2n-11-2=2n-1+3,当n=1时也适合,数列bn的前n项和为Tn=2n-1+3.
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