1、专题能力训练11等差数列与等比数列能力突破训练1.(2015全国高考)已知等比数列an满足a1=14,a3a5=4(a4-1),则a2=()A.2B.1C.12D.182.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.(2015浙江高考)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
2、.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.(2015全国高考)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.已知Sn为数列an的前n项和,且a2+S2=31,an+1
3、=3an-2n(nN*).(1)求证:an-2n为等比数列;(2)求数列an的前n项和Sn.10.已知数列an的前n项和为Sn,a1=1,an0,anan+1=Sn-1,其中为常数.(1)证明:an+2-an=;(2)是否存在,使得an为等差数列?并说明理由.11.已知数列an是等比数列.设a2=2,a5=16.(1)若a1+a2+a2n=t(a12+a22+an2),nN*,求实数t的值;(2)若在1a1与1a4之间插入k个数b1,b2,bk,使得1a1,b1,b2,bk,1a4,1a5成等差数列,求k的值.思维提升训练12.已知数列an,bn满足a1=b1=1,an+1-an=bn+1bn
4、=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.已知等比数列an的首项为43,公比为-13,其前n项和为Sn,若ASn-1SnB对nN*恒成立,则B-A的最小值为.15.等比数列an的各项均为正数,且2a1+3a2=1,a32=9a2a6.(1)求数列an的通项公式;(2)设bn=log3a1+log3a2+log3an,求数列1bn的前n项和
5、.16.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.参考答案能力突破训练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
6、+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,故q=13.由2a1+3a2=1得2a1+3a1q=1,所以a1=13.故数列an的通项公式为an=13n.
7、(2)bn=log3a1+log3a2+log3an=-(1+2+n)=-n(n+1)2.故1bn=-2n(n+1)=-21n-1n+1,1b1+1b2+1bn=-21-12+12-13+1n-1n+1=-2nn+1.所以数列1bn的前n项和为-2nn+1.16.解:(1)设等差数列an的公差为d,则d0,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+3.5