2020版高考數(shù)學(xué)復(fù)習(xí)第五單元專題探究4數(shù)列的綜合問題練習(xí)文（含解析）新人教A版.docx

專題探究4數(shù)列的綜合問題 1.2018南昌模擬 已知數(shù)列an為等比數(shù)列,a1=2,且a5是a3與a7的等差中項(xiàng),則a1008的值為()A.1或-1B.1C.2或-2D.22.2018廈門外國語學(xué)校月考 已知公差不為0的等差數(shù)列an滿足a32=a1a4,Sn為數(shù)列an的前n項(xiàng)和,則S3-S2S5-S3的值為()A.-2B.-3C.2D.33.2018鄂爾多斯一中模擬 如圖Z4-1,在等腰直角三角形ABC中,斜邊BC=22,過點(diǎn)A作BC的垂線,垂足為A1;過點(diǎn)A1作AC的垂線,垂足為A2;過點(diǎn)A2作A1C的垂線,垂足為A3以此類推,設(shè)BA=a1,AA1=a2,A1A2=a3,A5A6=a7,則a7=()圖Z4-1A.12B.1C.18D.144.2018成都模擬 設(shè)公比不為1的等比數(shù)列an滿足a1a2a3=-18,且a2,a4,a3成等差數(shù)列,則數(shù)列an的前4項(xiàng)和為.5.2018合肥模擬 已知數(shù)列an中,an=2n-1(nN*),則1a1a2+1a2a3+1a9a10=.6.2018安徽師大附中模擬 已知各項(xiàng)均為正數(shù)的數(shù)列an中,a1=1,前n項(xiàng)和Sn滿足SnSn-1-Sn-1Sn=2SnSn-1(nN*且n2),則a81=()A.641B.640C.639D.6387.2018信陽高級(jí)中學(xué)模擬 已知直線x+2y+5=0與直線x-dy+115=0互相平行且距離為m.若等差數(shù)列an的公差為d,且a7a8=35,a4+a100,令Sn=|a1|+|a2|+|a3|+|an|,則Sm的值為()A.60B.52C.44D.368.2018重慶模擬 已知數(shù)列an的通項(xiàng)公式為an=(-1)n(2n-1)cosn2+1(nN*),其前n項(xiàng)和為Sn,則S60=()A.-30B.-60C.90D.1209.2018吉安三校聯(lián)考 設(shè)Sn為等比數(shù)列an的前n項(xiàng)和,且關(guān)于x的方程a1x2-a3x+a2=0有兩個(gè)相等的實(shí)根,則S9S3=()A.27B.21C.14D.510.對(duì)于函數(shù)y=f(x),x與y的部分對(duì)應(yīng)關(guān)系如下表:x123456789y375961824數(shù)列xn滿足x1=1,且對(duì)于任意nN*,點(diǎn)(xn,xn+1)都在函數(shù)y=f(x)的圖像上,則x1+x2+x2015=()A.7554B.7549C.7546D.753911.2018佛山三水區(qū)實(shí)驗(yàn)中學(xué)月考 在等差數(shù)列an中,前10項(xiàng)和S10=120,則a3+a8的值是.12.2018安徽師大附中模擬 給出下面四個(gè)結(jié)論:若數(shù)列an的前n項(xiàng)和Sn=an2+bn+c(a,b,c為常數(shù)),則an為等差數(shù)列;若數(shù)列an是常數(shù)列,數(shù)列bn是等比數(shù)列,則數(shù)列anbn是等比數(shù)列;在等差數(shù)列an中,若公差d0,bn是首項(xiàng)為1的等差數(shù)列,a5+b3=21,a3+b5=13.(1)求數(shù)列an和bn的通項(xiàng)公式;(2)求數(shù)列bn2an的前n項(xiàng)和Sn.14.2018上海浦東新區(qū)三模 已知各項(xiàng)都不為零的無窮數(shù)列an滿足an+1an+an+1-an=0.(1)證明1an為等差數(shù)列,并求當(dāng)a1=1時(shí),數(shù)列an中的最大項(xiàng);(2)若a2018為數(shù)列an中的最小項(xiàng),求a1的取值范圍.15.2018株洲二模 已知數(shù)列an的前n項(xiàng)和為Sn,a1=1,且滿足anan+1=2Sn,數(shù)列bn滿足b1=15,bn+1-bn=2n,則數(shù)列bnan中第項(xiàng)最小.16.2018龍巖模擬 數(shù)列an是首項(xiàng)a10,公差為d的等差數(shù)列,其前n項(xiàng)和為Sn,若存在非零實(shí)數(shù)t,使得對(duì)任意nN*都有Sn=an+(n-1)tan恒成立,則t的值為.專題集訓(xùn)(四)1.C解析 由2a5=a3+a7q4-2q2+1=0q=1,所以a1008=a1q1007=2,故選C.2.C解析a32=a1a4,(a1+2d)2=a1(a1+3d),又d0,a1=-4d,則S3-S2S5-S3=a3a5+a4=a1+2d2a1+7d=-4d+2d-8d+7d=2.故選C.3.D解析 由題意得,a1=BA=2,a2=AA1=2,a3=A1A2=1,即數(shù)列an為等比數(shù)列,且首項(xiàng)為2,公比為22,則a7=2226=14.4.58解析 設(shè)等比數(shù)列an的公比為q(q1),由a1a2a3=-18,得a23=-18,解得a2=-12,則a3=-12q,a4=-12q2.因?yàn)閍2,a4,a3成等差數(shù)列,所以2a4=a2+a3,解得q=-12(q=1舍去),所以a1=1,所以數(shù)列an的前4項(xiàng)和S4=58.5.919解析 由an=2n-1,得1anan+1=1(2n-1)(2n+1)=1212n-1-12n+1,所以1a1a2+1a2a3+1a9a10=121-13+13-15+117-119=919.6.B解析 因?yàn)镾nSn-1-Sn-1Sn=2SnSn-1,所以Sn-Sn-1=2,即Sn為等差數(shù)列,且首項(xiàng)為1,公差為2,所以Sn=1+2(n-1)=2n-1,所以Sn=(2n-1)2,因此a81=S81-S80=1612-1592=640,故選B.7.B解析 由兩直線平行得d=-2,由兩平行直線間的距離公式得m=|115-5|1+22=10.a7(a7-2)=35,a7=-5或a7=7.又a4+a10=2a70時(shí),數(shù)列1an是正項(xiàng)遞增數(shù)列,此數(shù)列沒有最大項(xiàng),從而數(shù)列an中就沒有最小項(xiàng),故1a10.數(shù)列1an是遞增數(shù)列,且a2018是數(shù)列an中的最小項(xiàng),1a2018是數(shù)列1an中的最大負(fù)項(xiàng),從而有1a20180,即1a1+20170,-12017a10,即1a1+20180,a1-12018,a1的取值范圍是-12017,-12018.15.4解析 當(dāng)n2時(shí),2an=2Sn-2Sn-1=anan+1-an-1an,an0,an+1-an-1=2;當(dāng)n=1時(shí),a1a2=2a1,解得a2=2.數(shù)列an的奇數(shù)項(xiàng)與偶數(shù)項(xiàng)分別成等差數(shù)列,公差都為2,且a1=1,a2=2,進(jìn)而得到數(shù)列an為等差數(shù)列,其首項(xiàng)為1,公差為1,an=1+n-1=n.數(shù)列bn滿足b1=15,bn+1-bn=2n,當(dāng)n2時(shí),bn=2(n-1)+(n-2)+2+1+15=2(n-1)n2+15=n(n-1)+15,當(dāng)n=1時(shí),上式也成立,bn=n(n-1)+15,bnan=n-1+15n215-1(當(dāng)且僅當(dāng)n=15時(shí)取等號(hào)),又nN*,且b4a4=3+154,b3a3=7,則數(shù)列bnan中第4項(xiàng)最小.16.1或12解析 當(dāng)n=1時(shí),Sn=an+(n-1)tan恒成立.當(dāng)n2時(shí),當(dāng)數(shù)列an的公差d=0時(shí),Sn=an+(n-1)tan,即na1=a1+(n-1)ta1,即(n-1)a1=(n-1)ta1,則t=1;當(dāng)數(shù)列an的公差d0時(shí),由Sn=an+(n-1)tan,得Sn-1=an-1+(n-2)tan-1,