專題能力訓(xùn)練11等差數(shù)列與等比數(shù)列一、能力突破訓(xùn)練1.已知數(shù)列an為等比數(shù)列,且a8a9a10=-a132=-1 000,則a10a12=()A.100B.-100C.10010D.-100102.在等差數(shù)列an中,a1+a2+a3=3,a18+a19+a20=87,則此數(shù)列前20項(xiàng)的和等于()A.290B.300C.580D.6003.設(shè)an是等比數(shù)列,Sn是an的前n項(xiàng)和.若對(duì)任意正整數(shù)n,有an+2an+1+an+2=0,a1=2,則S101的值為()A.2B.200C.-2D.04.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>0B.a1d<0,dS4<0C.a1d>0,dS4<0D.a1d<0,dS4>05.在等比數(shù)列an中,滿足a1+a2+a3+a4+a5=3,a12+a22+a32+a42+a52=15,則a1-a2+a3-a4+a5的值是()A.3B.5C.-5D.56.在數(shù)列an中,a1=2,an+1=2an,Sn為an的前n項(xiàng)和.若Sn=126,則n=. 7.(2019四川內(nèi)江等六市二診,14)中國(guó)古代數(shù)學(xué)專著九章算術(shù)中有這樣一題:今有男子善走,日增等里,九日走1 260里,第一日、第四日、第七日所走之和為390里,則該男子第三日走的里數(shù)為. 8.設(shè)x,y,z是實(shí)數(shù),若9x,12y,15z成等比數(shù)列,且1x,1y,1z成等差數(shù)列,則xz+zx=. 9.(2018全國(guó),文17)在等比數(shù)列an中,a1=1,a5=4a3.(1)求an的通項(xiàng)公式;(2)記Sn為an的前n項(xiàng)和,若Sm=63,求m.10.(2019全國(guó),文18)記Sn為等差數(shù)列an的前n項(xiàng)和.已知S9=-a5.(1)若a3=4,求an的通項(xiàng)公式;(2)若a1>0,求使得Snan的n的取值范圍.11.(2019山東濰坊四市聯(lián)考,17)已知數(shù)列an,bn滿足:an+1+1=2an+n,bn-an=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.二、思維提升訓(xùn)練12.已知數(shù)列an,bn滿足a1=b1=1,an+1-an=bn+1bn=2,nN*,則數(shù)列ban的前10項(xiàng)的和為()A.43(49-1)B.43(410-1)C.13(49-1)D.13(410-1)13.若數(shù)列an為等比數(shù)列,且a1=1,q=2,則Tn=1a1a2+1a2a3+1anan+1等于()A.1-14nB.231-14nC.1-12nD.231-12n14.如圖,點(diǎn)列An,Bn分別在某銳角的兩邊上,且|AnAn+1|=|An+1An+2|,AnAn+2,nN*,|BnBn+1|=|Bn+1Bn+2|,BnBn+2,nN*.(PQ表示點(diǎn)P與Q不重合)若dn=|AnBn|,Sn為AnBnBn+1的面積,則()A.Sn是等差數(shù)列B.Sn2是等差數(shù)列C.dn是等差數(shù)列D.dn2是等差數(shù)列15.(2019河北武邑中學(xué)調(diào)研,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=5nn+5,則a10b9+b12+a11b8+b13=. 16.(2019江蘇常州高三期末,19)在數(shù)列an中,a1=1,且an+1+3an+4=0,nN*.(1)求證:an+1是等比數(shù)列,并求數(shù)列an的通項(xiàng)公式;(2)數(shù)列an中是否存在不同的三項(xiàng)按照一定順序重新排列后,構(gòu)成等差數(shù)列?若存在,求滿足條件的項(xiàng);若不存在,請(qǐng)說(shuō)明理由.17.若數(shù)列an是公差為正數(shù)的等差數(shù)列,且對(duì)任意nN*有an·Sn=2n3-n2.(1)求數(shù)列an的通項(xiàng)公式;(2)是否存在數(shù)列bn,使得數(shù)列anbn的前n項(xiàng)和為An=5+(2n-3)2n-1(nN*)?若存在,求出數(shù)列bn的通項(xiàng)公式及其前n項(xiàng)和Tn;若不存在,請(qǐng)說(shuō)明理由.專題能力訓(xùn)練11等差數(shù)列與等比數(shù)列一、能力突破訓(xùn)練1.C解析an為等比數(shù)列,a8a9a10=-a132=a93=-1000,a9=-10,a132=1000.又a10a12=a102q2>0,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解析設(shè)公比為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解析設(shè)an的首項(xiàng)為a1,公差為d,則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.d0,a1d=-53d2<0,且a1=-53d.dS4=4d(a1+a4)2=2d(2a1+3d)=-23d2<0,故選B.5.D解析由條件知a1(1-q5)1-q=3,a12(1-q10)1-q2=15,則a1(1+q5)1+q=5,故a1-a2+a3-a4+a5=a11-(-q)51-(-q)=a1(1+q5)1+q=5.6.6解析an+1=2an,即an+1an=2,an是以2為公比的等比數(shù)列.又a1=2,Sn=2(1-2n)1-2=126.2n=64,n=6.7.120解析男子每天走的里數(shù)構(gòu)成等差數(shù)列,設(shè)為an,其公差為d,前n項(xiàng)和為Sn.根據(jù)題意可知,S9=1260,a1+a4+a7=390,(方法一)S9=9(a1+a9)2=9a5=1260,a5=140.又a1+a4+a7=3a4=390,a4=130,d=a5-a4=10,a3=a4-d=120.(方法二)由題意,得S9=1260,a1+a4+a7=390,9a1+9×82d=1260,a1+a1+3d+a1+6d=390,解得a1=100,d=10,所以a3=a1+2d=120.8.3415解析由題意知(12y)2=9x×15z,2y=1x+1z,解得xz=1229×15y2=1615y2,x+z=3215y,從而xz+zx=x2+z2xz=(x+z)2-2xzxz=(x+z)2xz-2=32152y21615y2-2=3415.9.解(1)設(shè)an的公比為q,由題設(shè)得an=qn-1.由已知得q4=4q2,解得q=0(舍去)或q=-2或q=2.故an=(-2)n-1或an=2n-1.(2)若an=(-2)n-1,則Sn=1-(-2)n3.由Sm=63得(-2)m=-188,此方程沒(méi)有正整數(shù)解.若an=2n-1,則Sn=2n-1.由Sm=63得2m=64,解得m=6.綜上,m=6.10.解(1)設(shè)an的公差為d.由S9=-a5,得a1+4d=0.由a3=4,得a1+2d=4.于是a1=8,d=-2.因此an的通項(xiàng)公式為an=10-2n.(2)由(1)得a1=-4d,故an=(n-5)d,Sn=n(n-9)d2.由a1>0知d<0,故Snan等價(jià)于n2-11n+100,解得1n10.所以n的取值范圍是n|1n10,nN.11.解(1)因?yàn)閎n-an=n,所以bn=an+n.因?yàn)閍n+1=2an+n-1,所以an+1+(n+1)=2(an+n),即bn+1=2bn.又b1=2,所以bn是首項(xiàng)為2,公比為2的等比數(shù)列,bn=2×2n-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、思維提升訓(xùn)練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.則ban=2an-1=22(n-1)=4n-1,故數(shù)列ban的前10項(xiàng)和為1-4101-4=13(410-1).13.B解析因?yàn)閍n=1×2n-1=2n-1,所以anan+1=2n-1·2n=22n-1=2×4n-1,所以1anan+1=12×14n-1.所以1anan+1是等比數(shù)列.故Tn=1a1a2+1a2a3+1anan+1=12×1×1-14n1-14=231-14n.14.A解析如圖,延長(zhǎng)AnA1,BnB1交于P,過(guò)An作對(duì)邊BnBn+1的垂線,其長(zhǎng)度記為h1,過(guò)An+1作對(duì)邊Bn+1Bn+2的垂線,其長(zhǎng)度記為h2,則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).設(shè)此銳角為,則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均為定值,Sn+1-Sn為定值.Sn是等差數(shù)列.故選A.15.4解析由等差數(shù)列的性質(zhì)可得a10b9+b12+a11b8+b13=a10b1+b20+a11b1+b20=a1+a20b1+b20=20(a1+a20)220(b1+b20)2=S20T20=5×2020+5=4.16.解(1)因?yàn)閍n+1+3an+4=0,所以an+1+1an+1=-3an-3an+1=-3.因?yàn)閍1+1=20,所以數(shù)列an+1是以2為首項(xiàng),以-3為公比的等比數(shù)列,所以an+1=2×(-3)n-1,即an=2×(-3)n-1-1.(2)假設(shè)存在三項(xiàng)ar,as,at(r<s<t)按一定順序重新排列后成等差數(shù)列.若ar+as=2at,則2×(-3)r-1-1+2×(-3)s-1-1=4×(-3)t-1-2,整理得(-3)r+(-3)s=2×(-3)t.兩邊同除以(-3)r,得1+(-3)s-r=2×(-3)t-r,等式的右邊是-3的整數(shù)倍,左邊不是-3的整數(shù)倍,故等式不成立.若ar+at=2as,則2×(-3)r-1-1+2×(-3)t-1-1=4×(-3)s-1-2,整理得(-3)r+(-3)t=2×(-3)s,兩邊同除以(-3)r,得1+(-3)t-r=2×(-3)s-r,等式的右邊是-3的整數(shù)倍,左邊不是-3的整數(shù)倍,故等式不成立.若as+at=2ar,則2×(-3)s-1-1+2×(-3)t-1-1=4×(-3)r-1-2,整理得(-3)s+(-3)t=2×(-3)r,兩邊同除以(-3)r,可得(-3)s-r+(-3)t-r=2,等式的左邊是-3的整數(shù)倍,右邊不是-3的整數(shù)倍,故等式不成立.綜上,數(shù)列an中不存在不同的三項(xiàng)符合題意.17.解(1)設(shè)等差數(shù)列an的公差為d,則d>0,an=dn+(a1-d),Sn=12dn2+a1-12dn.對(duì)任意nN*,恒有an·Sn=2n3-n2,則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.d>0,a1=1,d=2,an=2n-1.(2)數(shù)列anbn的前n項(xiàng)和為An=5+(2n-3)·2n-1(nN*),當(dāng)n=1時(shí),a1b1=A1=4,b1=4,當(dāng)n2時(shí),anbn=An-An-1=5+(2n-3)2n-1-5+(2n-5)2n-2=(2n-1)2n-2.bn=2n-2.假設(shè)存在數(shù)列bn滿足題設(shè),且數(shù)列bn的通項(xiàng)公式bn=4,n=1,2n-2,n2,T1=4,當(dāng)n2時(shí),Tn=4+1-2n-11-2=2n-1+3,當(dāng)n=1時(shí)也適合,數(shù)列bn的前n項(xiàng)和為Tn=2n-1+3.10