2.3等比數(shù)列2.3.1等比數(shù)列【選題明細表】知識點、方法題號等比數(shù)列的定義及判定3,10,11等比數(shù)列的通項1,7,9等比數(shù)列的性質(zhì)2,4,6等差、等比數(shù)列的綜合5,81.已知{an}是等比數(shù)列,a2=2,a5=14(A)12 (B)2 (C)2 (D)解析:由a2=a1q,a5=a1q4,所以q3=a5a2=12.在等比數(shù)列{an}中,a1=1,a10=3,則a2a3(A)81 (B)275(C)3 (D)243解析:在等比數(shù)列{an}中,a1a10=a2a9=a3a8=a4a7=a5a6=3,則a23.(2017·貴州遵義市高一下期末)已知數(shù)列{an}前n項和Sn滿足:Sn=2an1(n∈N*),則該數(shù)列的第5項等于(B)(A)15 (B)16 (C)31 (D)32解析:根據(jù)題意,因為Sn=2an1,所以當(dāng)n=1時,a1=2a11,解得a1=1,當(dāng)n≥2時,an=SnSn1=(2an1)(2an11)=2an2an1,所以an=2an1,故an=2n1,則a5=251=16,故選B.4.正項等比數(shù)列{an}中,若log2(a2a98)=4,則a40a(A)4 (B)8 (C)16 (D)32解析:在等比數(shù)列中,a2a98=a40a所以由log2(a2a98得a2a98=24即a40a60故選C.5.(2017·北京卷)若等差數(shù)列{an}和等比數(shù)列{bn}滿足a1=b1=1,a4=b4=8,則a2b2=解析:設(shè)等差數(shù)列{an}的公差為d,等比數(shù)列{bn}的公比為q,依題意知,a4=a1+3d=1+3d=8,解得d=3.b4=b1q3=q3=8,解得q=2.所以a2=a1+d=1+3=2,b2=b1q=1×(2)=2,因此a2答案:16.在等比數(shù)列{an}中,若a7+a8+a9+a10=158,a8·a9=98,則1a7+1a8+1a9+解析:因為1a7+1a10=a7+a而a8a9=a7a所以1a7+1a8+1a9+1a答案:57.等比數(shù)列{an}中,a1+a2+a3+a4+a5=31,a2+a3+a4+a5+a6=62,則通項an等于(A)(A)2n1 (B)2n (C)2n+1 (D)2n+2解析:由a1(1+q+q2+q3+q4)=31①a2(1+q+q2+q3+q4)=62得q=2,代入①得a1=1,所以an=2n1.8.數(shù)列{an}為等差數(shù)列,a1,a2,a3成等比數(shù)列,a5=1,則a10等于(D)(A)5 (B)1 (C)0 (D)1解析:設(shè)公差為d,由已知得(解得a所以a10=a1+9d=1.故選D.9.已知各項不為0的等差數(shù)列{an}滿足a42a72+3a8=0,數(shù)列{bn}是等比數(shù)列,且b7=a7,則b2b8b11等于(D(A)1 (B)2 (C)4 (D)8解析:由已知,得2a72=a4+=a1+3d+3a1+21d=4a1+24d=4(a1+6d)=4a7,所以a7=2或a7=0(舍去),所以b7=2,所以b2b8b11=b1q·b1q7·b1q10=b13=(b1q6)3=b=8.10.已知a1=2,點(an,an+1)在函數(shù)f(x)=x2+2x的圖象上,其中n=1,2,3,…(1)證明數(shù)列{lg(1+an)}是等比數(shù)列;(2)求{an}的通項公式.(1)證明:由已知得an+1=an2+2a所以an+1+1=an2+2an+1=(an+1)因為a1=2,所以an+1+1=(an+1)2>0,所以lg(1+an+1)=2lg(1+an).即lg(1+an+1所以{lg(1+an)}是首項為lg3,公比為2的等比數(shù)列.(2)解:由(1)知,lg(1+an)=2n1·lg3=lg32所以1+an=32所以an=3211.數(shù)列{an}中,a1=2,a2=3,且{anan+1}是以3為公比的等比數(shù)列,記bn=a2n1+a2n(n∈N+).(1)求a3,a4,a5,a6的值;(2)求證:{bn}是等比數(shù)列.(1)解:因為{anan+1}是公比為3的等比數(shù)列,且a1·a2=6,所以anan+1=6·3n1=2·3n,所以a3=2×32a2a5=2×34a4(2)證明:由(1)得an·an+1=3an1·an(n≥2),所以an+1=3an1,所以a1,a3,a5,…,a2n1,…與a2,a4,a6,…,a2n,…都是公比