1、3等比数列3.1等比数列(一)基础过关1.在等比数列an中,满足2a4a6a5,则公比是()A.1 B.1或2C.1或2 D.1或2解析法一由已知得2a1q3a1q5a1q4,即2q2q,q1或q2.法二a5a4q,a6a4q2,由已知条件得2a4a4q2a4q,即2q2q,q1或q2.答案C2.下列数列为等比数列的是()A.2,22,222,B.,C.S1,(S1)2,(S1)3,D.0,0,0,解析A项中,A不是;B是首项为,公比为的等比数列;C项,当S1时,数列为0,0,0,不是;D显然不是.答案B3.已知等比数列an满足a1a23,a2a36,则a7等于()A.64 B.81 C.12
2、8 D.243解析q2代入a1a2a1(1q)3,得a11,a7a1q62664,故选A.答案A4.数列an满足:a91,an12an(nN),则a5_.解析由an12an(nN)知数列an是公比q2的等比数列.a5a1q4.答案5.在等比数列an中,a1a9256,a4a640,则公比q_.解析a1a9aq8,a4a6a1q3a1q5aq8,a1a9a4a6,列方程组解得或q2或q24.q或q2.答案2或2或或6.(1)一个等比数列an的第3项与第4项分别是12与18,求这个数列的通项公式;(2)已知等比数列an中,a53,a727,求q及an.解(1)法一设等比数列an的首项为a1,公比为
3、q,由题意得得ana1qn1.法二an为等比数列,q.ana3qn312.(2)由a7a5q2,得q29,q3,当q3时,ana5qn533n53n4;当q3时,ana5qn53(3)n5(3)n4.7.已知f(x)logmx(m0且m1),设f(a1),f(a2),f(an),是首项为4,公差为2的等差数列,求证:数列an是等比数列.证明由题意知f(an)42(n1)2n2logman,anm2n2,m2,m0且m1,m2为非零常数,数列an是等比数列.能力提升8.在等比数列an中,a3a44,a22,则公比q等于()A.2 B.1或2C.1 D.1或2解析根据题意,代入公式解得:或答案B9
4、.设a12,数列12an是公比为2的等比数列,则a6等于()A.31.5 B.160 C.79.5 D.159.5解析12a15,由题意知12an是以5为首项,公比为2的等比数列,所以12an52n1,所以an52n2,所以a652479.5.故选C.答案C10.若等差数列an和等比数列bn满足a1b11,a4b48,则_.解析an为等差数列,a11,a48a13d13d,d3,a2a1d132.bn为等比数列,b11,b48b1q3q3,q2,b2b1q2,则1.答案111.已知数列an为等差数列,其前n项和为Sn,S28,S432,数列bn为等比数列,且b1a1,b2(a2a1)b1,则b
5、n的通项公式为bn_.解析设公差为d,公比为q,由已知得所以又因为b2(a2a1)b1,所以q.又因为b1a12,所以bn2232n.答案232n12.数列an,bn满足下列条件,a10,a21,an2,bnan1an.(1)求证:bn是等比数列.(2)求bn的通项公式.(1)证明2an2anan1,bn是等比数列.(2)解b1a2a11,公比q,bn1.创新突破13.数列an满足a12,an1a6an6(nN),设cnlog5(an3).(1)求证:cn是等比数列;(2)求数列an的通项公式.(1)证明由an1a6an6,得an13(an3)2.log5(an13)log5(an3)22log5(an3),即cn12cn,又c1log5510,2,cn是等比数列.(2)解由(1)知,数列cn是以1为首项,以2为公比的等比数列,cn2n1,即log5(an3)2n1,an352n1.故an52n13.