1、第2课时等比数列的性质一、选择题1在等比数列an中,a2 0198a2 016,则公比q的值为()A2 B3 C4 D8答案A解析a2 0198a2 016a2 016q3,q38,q2.2在数列an中,a11,点(an,an1)在直线y2x上,则a4的值为()A7 B8 C9 D16答案B解析点(an,an1)在直线y2x上,an12an,a110,an0,an是首项为1,公比为2的等比数列,a41238.3已知等比数列an的公比为正数,且a5a74a,a21,则a1等于()A. B. C. D2答案B解析等比数列中,a5a7a,a4a.即(a4q2)24a.q44.q0,q.又a21,所以
2、a1.4在正项等比数列an中,a3a54,则a1a2a3a4a5a6a7等于()A64 B128 C256 D512答案B解析正项等比数列an中a1a7a2a6a3a5a4,a1a2a3a4a5a6a7a27128.5在正项等比数列an中,an1an,a2a86,a4a65,则等于()A. B. C. D.答案D解析设公比为q,则由等比数列an各项为正数且an1an知0q1的等比数列,若a4,a5是方程4x28x30的两根,则a6a7_.答案18解析由题意得a4,a5,q3.a6a7(a4a5)q23218.10已知等差数列an的公差为2,若a1,a3,a4成等比数列,则a2_.答案6解析由题
3、意知,a3a14,a4a16.a1,a3,a4成等比数列,aa1a4,(a14)2a1(a16),解得a18,a26.11已知等比数列an中,有a3a114a7,数列bn是等差数列,且b7a7,则b5b9_.答案8解析由等比数列的性质得a3a11a,a4a7.a70,a74,b7a74.再由等差数列的性质知b5b92b78.三、解答题12已知an为等比数列,且a1a964,a3a720,求a11.解因为an为等比数列,所以a1a9a3a764,又a3a720,所以a3,a7是方程t220t640的两个根所以a34,a716或a316,a74,当a34时,a3a7a3a3q420,所以1q45,
4、所以q44,所以a11a1q10a3q864;当a316时,a3a7a3(1q4)20,所以1q4,所以q4.所以a11a1q10a3q81.综上a1164或1.13有三个数成等比数列,其积为27,其平方和为91,求这三个数解设这三个数为,a,aq(公比为q),由已知得由得a3.将a3代入得q2,所以9q482q290,令q2t(t0),所以9t282t90,得t19,t2.所以q3或q.(1)当q3时,此数列为1,3,9;(2)当q3时,此数列为1,3,9;(3)当q时,此数列为9,3,1;(4)当q时,此数列为9,3,1.14已知等比数列an满足an0,且a5a2n522n(n3),则当n3时,log2a1log2a3log2a2n1等于()A2n B2n2 C2n2n Dn2答案D解析log2a1log2a3log2a2n1log2(a1a3a2n1) n2.15已知各项都为正数的数列an满足a11,a(2an11)an2an10.(1)求a2,a3;(2)求an的通项公式解(1)由题意可得1(2a21)2a20,解得a2,同理,可得a3.(2)由a(2an11)an2an10,得2an1(an1)an(an1)因为an的各项都为正数,所以.故an是首项为1,公比为的等比数列,因此an,nN.