1、开卷速查(三十)等比数列及其前n项和A级基础巩固练1已知等比数列an的公比q2,且2a4,a6,48成等差数列,则an的前8项和为()A127B255C511D.1 023解析:2a4,a6,48成等差数列,2a62a448,2a1q52a1q348,又q2,a11,S8255.答案:B2已知各项不为0的等差数列an满足a42a3a80,数列bn是等比数列,且b7a7,则b2b8b11等于()A1B.2C4D.8解析:a42a3a80,2aa43a8,即2a4a7,a72,b72,故b2b8b11b1qb1q7b1q10bq18(b7)38.答案:D3设等比数列an的前n项和为Sn,若Sm15
2、,Sm11,Sm121,则m()A3B.4C5D.6解析:由已知得,SmSm1am16,Sm1Smam132,故公比q2,又Sm11,故a11,又ama1qm116,故(1)(2)m116,求得m5.答案:C4已知等比数列an的前n项和为Sn,且a1a3,a2a4,则()A4n1B.4n1C2n1D.2n1解析:由除以可得2,解得q,代入得a12,an2n1.Sn4.2n1,选D.答案:D5等比数列an的各项均为正数,且a5a6a4a718,则log3a1log3a2log3a10()A12B.10C8D.2log35解析:由题意可知a5a6a4a7,又a5a6a4a718得a5a6a4a79
3、,而log3a1log3a2log3a10log3(a1a2a10)log3(a5a6)5log395log331010.答案:B6已知各项均为正数的等比数列an中,a4与a14的等比中项为2,则2a7a11的最小值为()A16B.8C2D.4解析:由题意知a40,a140,a4a148,a70,a110,则2a7a112228,当且仅当即a72,a114时取等号,故2a7a11的最小值为8,故选B.答案:B7在各项均为正数的等比数列an中,a12,a2a312,则该数列的前4项和为_解析:设等比数列an的公比为q,由a12,a2a312,则a1qa1q212,解得q2,故S430.答案:30
4、8若等比数列an满足a2a420,a3a540,则公比q_,前n项和Sn_.解析:由题意知q2.由a2a4a2(1q2)a1q(1q2)20,a12.Sn2n12.答案:22n129在等比数列an中,若a7a8a9a10,a8a9,则_.解析:,而a8a9a7a10,.答案:10已知公差不为0的等差数列an的前n项和为Sn,S3a46,且a1,a4,a13成等比数列(1)求数列an的通项公式;(2)设bn2an1,求数列bn的前n项和解析:(1)设等差数列an的公差为d(d0)由于S3a46,所以3a1a13d6.所以a13.由于a1,a4,a13成等比数列,所以a1(a112d)(a13d)
5、2,即3(312d)(33d)2.解得d2.所以an2n1.(2)由题意bn22n11,设数列bn的前n项和为Tn,cn22n1,4(nN*),所以数列cn为以8为首项,4为公比的等比数列所以Tnnn.B级力气提升练11已知数列an是首项为a1,公差为d(0d2)的等差数列,若数列cosan是等比数列,则其公比为()A1B.1C1D.2解析:由于数列cosan是等比数列,所以cos2(a1d)cosa1cos(a12d),cos2(a1d)cos(a1dd)cos(a1dd)cos2(a1d)cos2dsin2(a1d)sin2d,所以sin2dcos2(a1d)sin2(a1d)0,所以si
6、n2d0,sind0,由于0d2,所以d.公比q1.答案:B12已知等比数列an的公比为q,记bnam(n1)1am(n1)2am(n1)m,cnam(n1)1am(n1)2am(n1)m(m,nN*),则以下结论确定正确的是()A数列bn为等差数列,公差为qmB数列bn为等比数列,公比为q2mC数列cn为等比数列,公比为qm2D数列cn为等比数列,公比为qmm解析:an是等比数列,qmnmm(n1)mqm.(qm)mqm2.答案:C132021唐山市一中期中考试在数列an中,已知a11,an12ann1,nN*.(1)求证:ann是等比数列;(2)令bn,Sn为数列bn的前n项和,求Sn的表达式解析:(1)证明:由a11,an12ann1,nN*,可得an1(n1)2(ann),a1120,所以数列ann是以2为首项,以2为公比的等比数列(2)由(1)得:ann22n12n,所以ann2n,bn1,所以Snb1b2bnn.令Tn,则Tn,两式相减得Tn1所以Tn2,即Sn2n.142021贵州七校第一次联考已知an是等差数列,bn是等比数列,Sn为数列an的前n项和,a1b11,且b3S336,b2S28(nN*)(1)求an和bn;(2)若anan1,求数列的前n项和Tn.解析:(1)由题意得,解得或,或.(2)若anan1,由(1)知an2n1,Tn.