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.答案：C解析：∵a3a5=4(a4-1),∴=4(a4-1),解得a4=2.又a4=a1q3,且a1=,∴q=2,∴a2=a1q=.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.600答案：B解析：由a1+a2+a3=3,a18+a19+a20=87,得a1+a20=30,故S20==300.3.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.0答案：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===2.4.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.a1d>0,dS4>0B.a1d<0,dS4<0C.a1d>0,dS4<0D.a1d<0,dS4>0答案：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.∵d≠0,∴a1d=-d2<0,且a1=-d.∵dS4==2d(2a1+3d)=-d2<0,故选B.5.在等比数列{an}中,满足a1+a2+a3+a4+a5=3,++++=15,则a1-a2+a3-a4+a5的值是( )A.3B.C.-D.5答案：D解析：由条件知则=5,故a1-a2+a3-a4+a5===5.6.在数列{an}中,a1=2,an+1=2an,Sn为{an}的前n项和.若Sn=126,则n= . 答案：6解析：∵an+1=2an,即=2,∴{an}是以2为公比的等比数列.又a1=2,∴Sn==126.∴2n=64,∴n=6.7.已知等比数列{an}为递增数列,且=a10,2(an+an+2)=5an+1,则数列的通项公式an= . 答案：2n解析：∵=a10,∴(a1q4)2=a1q9,∴a1=q,∴an=qn.∵2(an+an+2)=5an+1,∴2an(1+q2)=5anq,4\n∴2(1+q2)=5q,解得q=2或q=(舍去),∴an=2n.8.设x,y,z是实数,若9x,12y,15z成等比数列,且,,成等差数列,则+= . 答案：解析：由题意知解得xz=y2=y2,x+z=y,从而+===-2=-2=.9.(2022北京,文15)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.解(1)设等差数列{an}的公差为d.因为a2+a4=10,所以2a1+4d=10.解得d=2.所以an=2n-1.(2)设等比数列{bn}的公比为q.因为b2b4=a5,所以b1qb1q3=9.解得q2=3.所以b2n-1=b1q2n-2=3n-1.从而b1+b3+b5+…+b2n-1=1+3+32+…+3n-1=.10.(2022全国Ⅲ,文17)设数列{an}满足a1+3a2+…+(2n-1)an=2n.(1)求{an}的通项公式;(2)求数列的前n项和.解(1)因为a1+3a2+…+(2n-1)an=2n,故当n≥2时,a1+3a2+…+(2n-3)an-1=2(n-1).两式相减得(2n-1)an=2.所以an=(n≥2).又由题设可得a1=2,从而{an}的通项公式为an=.(2)记的前n项和为Sn.由(1)知==-,则Sn=-+-+…+-=.11.(2022山东,文19)已知{an}是各项均为正数的等比数列,且a1+a2=6,a1a2=a3.(1)求数列{an}的通项公式;(2){bn}为各项非零的等差数列,其前n项和为Sn.已知S2n+1=bnbn+1,求数列的前n项和Tn.解(1)设{an}的公比为q,由题意知:a1(1+q)=6,q=a1q2,又an>0,解得a1=2,q=2,所以an=2n.(2)由题意知:S2n+1==(2n+1)bn+1,又S2n+1=bnbn+1,bn+1≠0,所以bn=2n+1.令cn=,则cn=,因此Tn=c1+c2+…+cn=+++…++.又Tn=+++…++,两式相减得Tn=+-,所以Tn=5-.二、思维提升训练12.已知数列{an},{bn}满足a1=b1=1,an+1-an==2,n∈N*,则数列{}的前10项的和为( )A.(49-1)B.(410-1)4\nC.(49-1)D.(410-1)答案：D解析：由a1=1,an+1-an=2,得an=2n-1.由=2,b1=1得bn=2n-1.则==22(n-1)=4n-1,故数列{}前10项和为=(410-1).13.若数列{an}为等比数列,且a1=1,q=2,则Tn=++…+等于( )A.1-B.C.1-D.答案：B解析：因为an=1×2n-1=2n-1,所以anan+1=2n-1·2n=22n-1=2×4n-1,所以=×.所以是等比数列.故Tn=++…+=×=.14.如图,点列{An},{Bn}分别在某锐角的两边上,且|AnAn+1|=|An+1An+2|,An≠An+2,n∈N*,|BnBn+1|=|Bn+1Bn+2|,Bn≠Bn+2,n∈N*.(P≠Q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.{}是等差数列C.{dn}是等差数列D.{}是等差数列答案：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=|BnBn+1|×h1,Sn+1=|Bn+1Bn+2|×h2.∴Sn+1-Sn=|Bn+1Bn+2|h2-|BnBn+1|h1.∵|BnBn+1|=|Bn+1Bn+2|,∴Sn+1-Sn=|BnBn+1|(h2-h1).设此锐角为θ,则h2=|PAn+1|sinθ,h1=|PAn|sinθ,∴h2-h1=sinθ(|PAn+1|-|PAn|)=|AnAn+1|sinθ.∴Sn+1-Sn=|BnBn+1||AnAn+1|sinθ.∵|BnBn+1|,|AnAn+1|,sinθ均为定值,∴Sn+1-Sn为定值.∴{Sn}是等差数列.故选A.15.已知等比数列{an}的首项为,公比为-,其前n项和为Sn,若A≤Sn-≤B对n∈N*恒成立,则B-A的最小值为 . 答案：解析：易得Sn=1-∈∪,4\n因为y=Sn-在区间上单调递增(y≠0),所以y∈⊆[A,B],因此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q>0,n∈N*.(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-=1的离心率为en,且e2=2,求++…+.解(1)由已知,Sn+1=qSn+1,Sn+2=qSn+1+1,两式相减得到an+2=qan+1,n≥1.又由S2=qS1+1得到a2=qa1,故an+1=qan对所有n≥1都成立.所以,数列{an}是首项为1,公比为q的等比数列.从而an=qn-1.由a2,a3,a2+a3成等差数列,可得2a3=a2+a2+a3.所以a3=2a2,故q=2.所以an=2n-1(n∈N*).(2)由(1)可知,an=qn-1.所以双曲线x2-=1的离心率en==.由e2==2,解得q=.所以++…+=(1+1)+(1+q2)+…+[1+q2(n-1)]=n+[1+q2+…+q2(n-1)]=n+=n+(3n-1).17.若数列{an}是公差为正数的等差数列,且对任意n∈N*有an·Sn=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(n∈N*)?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.解(1)设等差数列{an}的公差为d,则d>0,an=dn+(a1-d),Sn=dn2+n.对任意n∈N*,恒有an·Sn=2n3-n2,则[dn+(a1-d)]·=2n3-n2,即[dn+(a1-d)]·=2n2-n.∴∵d>0,∴∴an=2n-1.(2)∵数列{anbn}的前n项和为An=5+(2n-3)·2n-1(n∈N*),∴当n=1时,a1b1=A1=4,∴b1=4,当n≥2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=∴T1=4,当n≥2时,Tn=4+=2n-1+3,当n=1时也适合,∴数列{bn}的前n项和为Tn=2n-1+3.4
