1、学业分层测评(十二)(建议用时:45分钟)学业达标一、选择题12与2的等比中项是()A1B1C1D2【解析】2与2的等比中项为G1,故选C.【答案】C2在等比数列an中,a2 0168a2 015,则公比q的值为()A2B3C4D8【解析】因为a2 0168a2 015,所以a1q2 0158a1q2 014,解得q8.【答案】D3已知一等比数列的前三项依次为x,2x2,3x3,那么13是此数列的()A第2项B第4项C第6项D第8项【解析】由x,2x2,3x3成等比数列,可知(2x2)2x(3x3),解得x1或4,又当x1时,2x20,这与等比数列的定义相矛盾x4,该数列是首项为4,公比为的等
2、比数列,其通项an4n1,由4n113,得n4.【答案】B4已知a,b,c,d成等比数列,且曲线yx22x3的顶点坐标是(b,c),则ad等于()A3B2C1D2【解析】由yx22x3(x1)22,得b1,c2.又a,b,c,d成等比数列,即a,1,2,d成等比数列,所以d4,a,故ad42.【答案】B5(2015全国卷)已知等比数列an满足a13,a1a3a521,则a3a5a7()A21B42C63D84【解析】a13,a1a3a521,33q23q421,1q2q47,解得q22或q23(舍去)a3a5a7q2(a1a3a5)22142.故选B.【答案】B二、填空题6已知等比数列an中,
3、a12,且a4a64a,则a3 .【解析】设等比数列an的公比为q,由已知条件得a4aq4.q4,q2,a3a1q221.【答案】17已知等比数列an中,a33,a10384,则该数列的通项an .【解析】由已知得q712827,故q2.所以ana1qn1a1q2qn3a3qn332n3.【答案】32n38在等比数列an中,an0,且a1a21,a3a49,则a4a5 .【解析】由已知a1a21,a3a49,q29.q3(q3舍),a4a5(a3a4)q27.【答案】27三、解答题9在各项均为负的等比数列an中,2an3an1,且a2a5.(1)求数列an的通项公式;(2)是否为该数列的项?若
4、是,为第几项?【解】(1)因为2an3an1,所以,数列an是公比为的等比数列,又a2a5,所以a53,由于各项均为负,故a1,ann2.(2)设an,则n2,n24,n6,所以是该数列的项,为第6项10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【解】(1)证明:2an2anan1,.bn是等比数列(2)b1a2a11,公比q,bn1n1n1.能力提升1已知等比数列an中,各项都是正数,且a1,a3,2a2成等差数列,则等于()A.1B32C32D23【解析】设等比数列an的公比为q,由于a1,a3,2a2成等差数列,
5、则2a12a2,即a3a12a2,所以a1q2a12a1q.由于a10,所以q212q,解得 q1.又等比数列an中各项都是正数,所以q0,所以q1.所以32.【答案】322(2015全国卷)已知等比数列an满足a1,a3a54(a41),则a2()A2B1C.D【解析】法一a3a5a,a3a54(a41),a4(a41),a4a440,a42.又q38,q2,a2a1q2,故选C.法二a3a54(a41),a1q2a1q44(a1q31),将a1代入上式并整理,得q616q3640,解得q2,a2a1q,故选C.【答案】C3(2015浙江高考)已知an是等差数列,公差d不为零若a2,a3,a7成等比数列,且2a1a21,则a1 ,d .【解析】a2,a3,a7成等比数列,aa2a7,(a12d)2(a1d)(a16d),即2d3a10.又2a1a21,3a1d1.由解得a1,d1.【答案】14已知数列an满足a11,an12an1.(1)求证:数列an1是等比数列;(2)求an的表达式. 【导学号:05920070】【解】(1)证明:an12an1,an112(an1)由a11,故a110,由上式易知an10,2.an1是等比数列(2)由(1)可知an1是以a112为首项,以2为公比的等比数列,an122n1,即an2n1.