新教材2024-2025學年高中數(shù)學第一章數(shù)列3等比數(shù)列3.1等比數(shù)列第2課時等比數(shù)列的性質(zhì)及應用分層作業(yè)北師大版選擇性必修第二冊

第一章第2課時等比數(shù)列的性質(zhì)及應用A級必備學問基礎練1.(多選題)已知{an}是等比數(shù)列,a2=2,a6=18,則公比q=(A.-12 B.-2 C.2 D.2.在等比數(shù)列{an}中,若a2a6+a42=π,則a3a5等于 (A.π4 B.π3 C.π3.(多選題)已知等比數(shù)列{an},a1=1,q=2,則()A.數(shù)列1aB.數(shù)列1aC.數(shù)列{log2an}是等差數(shù)列D.數(shù)列{log2an}是遞增數(shù)列4.(多選題)[2024遼寧鞍山一中校考期中]已知數(shù)列{an},{bn},下列說法正確的有()A.若an=2-5n,則{an}為遞減數(shù)列B.若b1≠0,bn+1=3bn,則{bn}為等比數(shù)列C.若數(shù)列{bn}的公比q=-1,則{bn}為遞減數(shù)列D.若數(shù)列{an}的前n項和Sn=n2+2n,則{an}為等差數(shù)列5.(多選題)[2024黑龍江哈爾濱三中階段練習]已知等差數(shù)列{an}的公差d和首項a1都不等于0,且a3,a5,a8成等比數(shù)列,則下列說法正確的是()A.a1+a5C.a1=2d D.d=2a16.公比不為1的等比數(shù)列{an}滿意a5a6+a4a7=18,若a1am=9,則m的值為()A.8 B.9 C.10 D.117.已知等差數(shù)列{an}的公差為2,若a1,a3,a4成等比數(shù)列,則a2=.

8.在1與2之間插入6個正數(shù),使這8個數(shù)成等比數(shù)列,則插入的6個數(shù)的積為.

9.已知在等比數(shù)列{an}中,有a3a11=4a7,數(shù)列{bn}是等差數(shù)列,且b7=a7,則b5+b9=.

10.已知數(shù)列{an}是等差數(shù)列,Sn為其前n項和,a4=9,S4=24.(1)求{an}的通項公式;(2)若bn=3an,求證:{bn11.已知數(shù)列{an}是等比數(shù)列,a3+a7=20,a1a9=64,求a11的值.B級關鍵實力提升練12.已知等比數(shù)列{an}共有10項,其中奇數(shù)項之積為2,偶數(shù)項之積為64,則其公比是()A.32 B.2 C.2 D.213.已知在各項均為正數(shù)的等比數(shù)列{an}中,lg(a3a8a13)=6,則a1a15的值為()A.100 B.-100C.10000 D.-1000014.在正項等比數(shù)列{an}中,已知a1a2a3=4,a4a5a6=12,an-1anan+1=324,則n等于()A.12 B.13 C.14 D.1515.(多選題)[2024北京高二階段練習]已知數(shù)列{an}是公差不為0的等差數(shù)列,其前n項和為Sn,且a1,a2,a5成等比數(shù)列,則下列說法正確的是()A.若a1>0,則S4>a8 B.若a1>0,則S4≤a8C.若a1<0,則S4>a8 D.若a1<0,則S4<a816.(多選題)[2024吉林白城高二階段練習]若數(shù)列{an}為等比數(shù)列,則下列肯定成立的是()A.若a3>0,則a2023<0 B.若a4>0,則a2022<0C.若a3>0,則a2021>0 D.若a4>0,則a2020>017.(多選題)[2024貴州黔西南高二統(tǒng)考期末]若等比數(shù)列{an}的第4項和第6項分別是48和12,下列選項中說法正確的是()A.{an}的公比為12或-B.{an}的第5項是24C.a3·a2022=a1·a2024D.a3+a2022=a1+a202418.設數(shù)列{an}為公比q>1的等比數(shù)列,若a4,a5是方程4x2-8x+3=0的兩根,則a6+a7=.

19.在等比數(shù)列{an}中,若a1a2a3a4=1,a13a14a15a16=8,則a41a42a43a44=.

20.已知等比數(shù)列{an}的各項均為正數(shù),且a6=2,a4+a5=12.(1)求數(shù)列{an}的通項公式;(2)設bn=a1a3a5…a2n-1,n∈N+,求數(shù)列{bn}的最大項.C級學科素養(yǎng)創(chuàng)新練21.記Sn為數(shù)列{an}的前n項和,已知an>0,a2=2a1,且數(shù)列{Sn+a1}是等比數(shù)列,求證:{an}是等比數(shù)列.參考答案第2課時等比數(shù)列的性質(zhì)及應用1.AD由題意可得q4=a6a2=116,解得故選AD.2.Ca2a6=a42=a3a5,∴a3a5=3.ACD由a1=1,q=2得an=2n-1,1an=12n-1,所以數(shù)列1an是等比數(shù)列且為遞減數(shù)列,故A正確,B不正確;log2a4.ABD對于A,當n∈N+時,an+1-an=2-5(n+1)-(2-5n)=-5<0,即an+1<an,A正確;對于B,因為b1≠0,n∈N+,所以bn≠0,由已知得bn+1bn=對于C,當b1=1時,b2=-b1=-1,b3=-b2=1,則b3>b2,故{bn}不是遞減數(shù)列,C錯誤;對于D,由Sn=n2+2n得當n=1時,a1=S1=3,當n≥2時,an=Sn-Sn-1=n2+2n-(n-1)2-2n+2=2n+1,檢驗得,當n=1時,滿意an=2n+1,所以an+1-an=2,則{an}為等差數(shù)列,D正確.故選ABD.5.BC由a3,a5,a8成等比數(shù)列,則a3a8=a52,即(a1+2d)(a1+7d)=(a1+4d)2,即a1d=2d2,又因為公差d和首項a1都不等于0,則a1=2所以a1+a故選BC.6.C由題意得,2a5a6=18,a5a6=9,∵a1am=9,∴a1am=a5a6,∴m=10,故選C.7.-6由題意知,a3=a1+4,a4=a1+6.∵a1,a3,a4成等比數(shù)列,∴a32=a1a∴(a1+4)2=(a1+6)a1,解得a1=-8,∴a2=-6.8.8設這8個數(shù)組成的等比數(shù)列為{an},則a1=1,a8=2.插入的6個數(shù)的積為a2a3a4a5a6a7=(a2a7)·(a3a6)·(a4a5)=(a1a8)3=23=8.9.8由等比數(shù)列的性質(zhì),得a3a11=a72,∴a72=∵a7≠0,∴a7=4,∴b7=a7=4.再由等差數(shù)列的性質(zhì)知b5+b9=2b7=8.10.(1)解因為{an}是等差數(shù)列,a4=9,S4=24,所以a1+3所以an=3+2(n-1)=2n+1.(2)證明bn=3an=32n+1,所以b所以{bn}是以9為公比的等比數(shù)列.11.解∵{an}是等比數(shù)列,∴a1·a9=a3·a7=64.又a3+a7=20,∴a3=4,a7=16或a3=16,a7=4.①當a3=4,a7=16時,a7a3=q4=4,此時a11=a3q8=4×42②當a3=16,a7=4時,a7a3=q4=14,此時a11=a3q8=16×14212.C∵奇數(shù)項之積為2,偶數(shù)項之積為64,∴a1a3a5a7a9=2,a2a4a6a8a10=64,則a2a4a6a813.C∵lg(a3a8a13)=lga83∴a83=106,∴a8=102=100.∴a1a15=a814.C設數(shù)列{an}的公比為q,由a1a2a3=4=a13q3與a4a5a6=12=a13q12,可得q9=3,an-1anan+1=a13q3n-3=324,因此q3n-6=81=3415.AD設等差數(shù)列{an}的公差為d≠0,∵a1,a2,a5成等比數(shù)列,則a22=a1a可得(a1+d)2=a1(a1+4d),整理得d2=2a1d,由d≠0,則d=2a1,則S4-a8=(4a1+6d)-(a1+7d)=3a1-d=3a1-2a1=a1,對于A,B,若a1>0,即S4-a8=a1>0,故S4>a8,A正確,B錯誤;對于C,D,若a1<0,即S4-a8=a1<0,故S4<a8,D正確,C錯誤.故選AD.16.CD若a3=a1q2>0,則a1>0,所以a2024=a1q2024>0,故A錯誤;若a4=a1q3>0,則a2024=a1q2024=a1q3·q2024>0,故B錯誤;若a3=a1q2>0,則a1>0,所以a2024=a1q2024>0,故C正確;若a4=a1q3>0,則a2024=a1q2024=a1q3·q2016>0,故D正確.故選CD.17.AC設該等比數(shù)列的公比為q,由題意可知,a6=a4q2,則q2=1248=14,解得a5=a4q=48×±12由等比數(shù)列性質(zhì)知,a3·a2024=a1·a2024,故選項C正確,D不正確.故選AC.18.18由題意得a4=12,a5=32,∴q=a5∴a6+a7=(a4+a5)q2=12+32×32=19.1024設等比數(shù)列{an}的公比為q,a1a2a3a4=a1·a1q·a1q2·a1q3=a14·q6=a13a14a15a16=a1q12·a1q13·a1q14·a1q15=a14·q54=②÷①得q48=8,q16=2,∴a41a42a43a44=a1q40·a1q41·a1q42·a1q43=a14·q166=a14·q6·q160=(a14·q6)(q16)1020.解(1)設等比數(shù)列{an}的公比為q(q>0),由a6=2,a4+a5=12,得a解得a1∴an=64×12n-1=(2)bn=a1a3a5…a2n