（課標(biāo)專用）天津市2020高考數(shù)學(xué)二輪復(fù)習(xí) 專題能力訓(xùn)練10 等差數(shù)列與等比數(shù)列

專題能力訓(xùn)練10等差數(shù)列與等比數(shù)列專題能力訓(xùn)練第26頁 一、能力突破訓(xùn)練1.在等差數(shù)列an中,a4+a10+a16=30,則a18-2a14的值為()A.20B.-20C.10D.-10答案:D解析:因?yàn)閍4+a10+a16=30,所以3a10=30,即a10=10,所以a18-2a14=-a10=-10.故選D.2.在各項(xiàng)均為正數(shù)的等比數(shù)列an中,若log2(a2·a3·a5·a7·a8)=5,則a1·a9=()A.4B.5C.2D.25答案:A解析:由題意得log2(a2·a3·a5·a7·a8)=log2a55=5log2a5=5,所以a5=2.所以a1·a9=a52=4.故選A.3.設(shè)an是等比數(shù)列,Sn是an的前n項(xiàng)和.對任意正整數(shù)n,有an+2an+1+an+2=0,又a1=2,則S101的值為()A.2B.200C.-2D.0答案: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.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>0答案: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.中國古代數(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. 答案:120解析:由男子善走,日增等里,可知每天走的里數(shù)符合等差數(shù)列,設(shè)這個(gè)等差數(shù)列為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.a1+a4+a7=3a4=390,a4=130,d=a5-a4=10,a3=a4-d=120.(方法二)S9=1260,a1+a4+a7=390,即9a1+9×82d=1260,a1+a1+3d+a1+6d=390,解得a1=100,d=10,a3=a1+2d=120.6.已知各項(xiàng)均為正數(shù)的等差數(shù)列an的前n項(xiàng)和為Sn,S10=40,則a3·a8的最大值為. 答案:16解析:因?yàn)镾10=10(a1+a10)2=40a1+a10=a3+a8=8,a3>0,a8>0,所以a3·a8a3+a822=822=16,當(dāng)且僅當(dāng)a3=a8=4時(shí)取等號.7.設(shè)等比數(shù)列an滿足a1+a3=10,a2+a4=5,則a1a2an的最大值為. 答案:64解析:由已知a1+a3=10,a2+a4=a1q+a3q=5,兩式相除得a1+a3q(a1+a3)=105,解得q=12,a1=8,所以a1a2an=8n·121+2+(n-1)=2-12n2+7n2,拋物線f(n)=-12n2+72n的對稱軸為直線n=-722×-12=3.5,又nN*,所以當(dāng)n=3或4時(shí),a1a2an取最大值為2-12×32+7×32=26=64.8.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=. 答案: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.已知Sn為數(shù)列an的前n項(xiàng)和,且a2+S2=31,an+1=3an-2n(nN*).(1)求證:an-2n為等比數(shù)列;(2)求數(shù)列an的前n項(xiàng)和Sn.(1)證明由an+1=3an-2n可得an+1-2n+1=3an-2n-2n+1=3an-3·2n=3(an-2n).又a2=3a1-2,則S2=a1+a2=4a1-2,得a2+S2=7a1-4=31,得a1=5,則a1-21=30.故an-2n為等比數(shù)列.(2)解由(1)可知an-2n=3n-1(a1-2)=3n,an=2n+3n,Sn=2(1-2n)1-2+3(1-3n)1-3=2n+1+3n+12-72.10.記Sn為等差數(shù)列an的前n項(xiàng)和,已知a1=-7,S3=-15.(1)求an的通項(xiàng)公式;(2)求Sn,并求Sn的最小值.解:(1)設(shè)an的公差為d,由題意得3a1+3d=-15.由a1=-7得d=2.所以an的通項(xiàng)公式為an=2n-9.(2)由(1)得Sn=n2-8n=(n-4)2-16.所以當(dāng)n=4時(shí),Sn取得最小值,最小值為-16.11.已知數(shù)列an是等比數(shù)列.設(shè)a2=2,a5=16.(1)若a1+a2+a2n=t(a12+a22+an2),nN*,求實(shí)數(shù)t的值;(2)若在1a1與1a4之間插入k個(gè)數(shù)b1,b2,bk,使得1a1,b1,b2,bk,1a4,1a5成等差數(shù)列,求k的值.解:設(shè)等比數(shù)列an的公比為q,由a2=2,a5=16,得q=2,a1=1.(1)a1+a2+a2n=t(a12+a22+an2),a1(1-q2n)1-q=t·a12(1-q2n)1-q2,即1-22n1-2=t·1-22n1-4對nN*都成立,t=3.(2)1a1=1,1a4=18,1a5=116,且1a1,b1,b2,bk,1a4,1a5成等差數(shù)列,公差d=1a5-1a4=-116,且1a4-1a1=(k+1)d,即18-1=(k+1)×-116,解得k=13.二、思維提升訓(xùn)練12.幾位大學(xué)生響應(yīng)國家的創(chuàng)業(yè)號召,開發(fā)了一款應(yīng)用軟件.為激發(fā)大家學(xué)習(xí)數(shù)學(xué)的興趣,他們推出了“解數(shù)學(xué)題獲取軟件激活碼”的活動.這款軟件的激活碼為下面數(shù)學(xué)問題的答案已知數(shù)列1,1,2,1,2,4,1,2,4,8,1,2,4,8,16,其中第一項(xiàng)是20,接下來的兩項(xiàng)是20,21,再接下來的三項(xiàng)是20,21,22,依此類推.求滿足如下條件的最小整數(shù)N:N>100且該數(shù)列的前N項(xiàng)和為2的整數(shù)冪.那么該款軟件的激活碼是()A.440B.330C.220D.110答案:A解析:設(shè)數(shù)列的首項(xiàng)為第1組,接下來兩項(xiàng)為第2組,再接下來三項(xiàng)為第3組,以此類推,設(shè)第n組的項(xiàng)數(shù)為n,則前n組的項(xiàng)數(shù)和為n(1+n)2.第n組的和為1-2n1-2=2n-1,前n組總共的和為2(1-2n)1-2-n=2n+1-2-n.由題意,N>100,令n(1+n)2>100,得n14且nN*,即N出現(xiàn)在第13組之后.若要使最小整數(shù)N滿足:N>100且前N項(xiàng)和為2的整數(shù)冪,則SN-Sn(1+n)2應(yīng)與-2-n互為相反數(shù),即2k-1=2+n(kN*,n14),所以k=log2(n+3),解得n=29,k=5.所以N=29×(1+29)2+5=440,故選A.13.設(shè)等比數(shù)列an的前n項(xiàng)和為Sn,若8a2+a5=0,則下列式子中數(shù)值不能確定的是()A.a5a3B.S5S3C.an+1anD.Sn+1Sn答案:D解析:等比數(shù)列an中,8a2+a5=0,a5=-8a2,q=-2,a5a3=4,an+1an=q=-2,S5S3=a11-(-2)51+2a11-(-2)31+2=113,Sn+1Sn不能確定.14.若存在等比數(shù)列an,使得a1(a2+a3)=6a1-9,則公比q的取值范圍為. 答案:-1-52,00,-1+52解析:因?yàn)閍2+a3=a1(q+q2),所以a12(q+q2)-6a1+9=0.當(dāng)q+q2=0時(shí),易知q=-1滿足題意,但q0;當(dāng)q+q20時(shí),=36-36(q+q2)0,解得-1-52q-1+52,綜上,q-1-52,00,-1+52.15.設(shè)Sn為等差數(shù)列an的前n項(xiàng)和,a2+a3=8,S9=81.(1)求an的通項(xiàng)公式;(2)若S3,a14,Sm成等比數(shù)列,求S2m.解:(1)S9=9a5=9(a1+4d)=81,a2+a3=2a1+3d=8,a1=1,d=2,故an=1+(n-1)×2=2n-1.(2)由(1)知,Sn=n(1+2n-1)2=n2.S3,a14,Sm成等比數(shù)列,S3·Sm=a142,即9m2=272,解得m=9,故S2m=182=324.16.等比數(shù)列an的各項(xiàng)均為正數(shù),且2a1+3a2=1,a32=9a2a6.(1)求數(shù)列an的通項(xiàng)公式;(2)設(shè)bn=log3a1+log3a2+log3an,求數(shù)列1bn的前n項(xiàng)和.解:(1)設(shè)數(shù)列an的公比為q.由a32=9a2a6得a32=9a42,所以q2=19.由條件可知q>0,故q=13.由2a1+3a2=1得2a1+3a1q=1,所以a1=13.故數(shù)列an的通項(xiàng)公式為an=13n.(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.所以數(shù)列1bn的前n項(xiàng)和為-2nn+1.17.若數(shù)列an是公差為正數(shù)的等差數(shù)列,且對任意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,請說明理由.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+3.7