新教材適用2023-2024學(xué)年高中數(shù)學(xué)第一章數(shù)列3等比數(shù)列3.1等比數(shù)列第2課時(shí)等比中項(xiàng)及等比數(shù)列的性質(zhì)課后訓(xùn)練北師大版選擇性必修第二冊(cè)

第2課時(shí)等比中項(xiàng)及等比數(shù)列的性質(zhì)課后訓(xùn)練鞏固提升A組1.在正項(xiàng)等比數(shù)列{an}中,若log2(a2a98)=4,則a40a60等于().A.4 B.8 C.16 D.32解析:在等比數(shù)列{an}中,a2a98=a40a60,所以由log2(a2a98)=4,得a2a98=24=16,即a40a60=16.故選C.答案:C2.在等比數(shù)列{an}中,a1=8,a4=a3a5,則a7=().A.116 B.1C.14 D.解析:在等比數(shù)列{an}中,a42=a3a又a4=a3a5,所以a4=1,故q=12,所以a7=1答案:B3.在各項(xiàng)均為正數(shù)的等比數(shù)列{an}中,a3=2-1,a5=2+1,則a32+2a2a6+a3a7=(A.4 B.6 C.8 D.8-42解析:a3+a5=2-1+2+1=22,故a32+2a2a6+a3a7=a32+2a3a5+a52=(a3+a5)答案:C4.在等比數(shù)列{an}中,a5·a11=3,a3+a13=4,則a12a2=(A.3 B.-13C.3或13 D.-3或-解析:因?yàn)樵诘缺葦?shù)列{an}中,a5·a11=3,a3+a13=4,所以由等比數(shù)列的性質(zhì)得,(化簡(jiǎn)得,3q20-10q10+3=0,解得q10=3或q10=13即a12a2=q10=3,或a12答案:C5.已知1是a2與b2的等比中項(xiàng),又是1a與1b的等差中項(xiàng),則a+A.1或12 B.1或-1C.1或13 D.1或-解析:∵a2b2=1,1a+1b=2,∴a2+2∴a2+b2=4-2ab,又a+b=2ab,∴a+由a2b2=1知ab=±1,∴a+ba2+b答案:D6.在等比數(shù)列{an}中,若a1·a2·a3·a4=1,a13·a14·a15·a16=8,則a41·a42·a43·a44=.

解析:a1·a2·a3·a4=a1·a1q·a1q2·a1q3=a14·q6=a13·a14·a15·a16=a1q12·a1q13·a1q14·a1q15=a14·q54=由②÷①,得a14·q54a14·q又a41·a42·a43·a44=a1q40·a1q41·a1q42·a1q43=a14·q166=a14·q6·q160=(a14·q6)·(q16)10=1答案:10247.在等比數(shù)列{an}中,若a3,a7是方程x2+4x+2=0的兩根,則a3a7=;a5的值是.

解析:根據(jù)根與系數(shù)的關(guān)系得a3+a7=-4,a3a7=2,由a3+a7=-4<0,a3a7>0,得a3<0,a7<0,故a5<0.由a3a7=a52,得a5=-a3答案:2-28.三個(gè)互不相等的數(shù)成等差數(shù)列,適當(dāng)排列這三個(gè)數(shù)后,又可成為等比數(shù)列,已知這三個(gè)數(shù)的和為6,求這三個(gè)數(shù).解:設(shè)這三個(gè)數(shù)分別為a-d,a,a+d,則a-d+a+a+d=6,得a=2,即這三個(gè)數(shù)為2-d,2,2+d.①若2-d為等比中項(xiàng),則有(2-d)2=2(2+d),解得d=6或d=0(舍).此時(shí)三個(gè)數(shù)為-4,2,8.②若2+d為等比中項(xiàng),則有(2+d)2=2(2-d),解得d=-6或d=0(舍).此時(shí)三個(gè)數(shù)為8,2,-4.③若2為等比中項(xiàng),則有22=(2-d)·(2+d),解得d=0(舍).綜上,此三個(gè)數(shù)為-4,2,8.9.已知數(shù)列{an}的前n項(xiàng)和Sn=3n2-n2,(1)求數(shù)列{an}的通項(xiàng)公式;(2)證明:對(duì)任意的n>1,都存在m∈N+,使得a1,an,am成等比數(shù)列.(1)解:由Sn=3n2-n2,得a1=S1=1,當(dāng)n≥2時(shí),an=Sn-Sn-1=3n-2,將a所以數(shù)列{an}的通項(xiàng)公式為an=3n-2.(2)證明:要使得a1,an,am成等比數(shù)列,只需要an2=a1·am,即(3n-2)2=1·(3m-2),即m=3n2-4n+2,而此時(shí)m∈N+,且所以對(duì)任意的n>1,都存在m∈N+,使得a1,an,am成等比數(shù)列.B組1.已知等比數(shù)列{an}的各項(xiàng)都是正數(shù),且a1,12a3,2a2成等差數(shù)列,則a6+a7A.2+1 B.3+22 C.3-22 D.22-3解析:設(shè)等比數(shù)列{an}的公比為q,由于a1,12a3,2a2成等差數(shù)列則212a3=a1+2a2,即a3=a1+2所以a1q2=a1+2a1q.因?yàn)閍1≠0,所以q2=1+2q,解得q=1±2.又等比數(shù)列{an}中各項(xiàng)都是正數(shù),所以q>0,所以q=1+2.所以a6+a7a8答案:C2.已知等比數(shù)列{an}滿足an>0,n=1,2,3,…,且a5·a2n-5=22n(n≥3),則當(dāng)n≥1時(shí),log2a1+log2a3+…+log2a2n-1=().A.n(2n-1) B.(n+1)2 C.n2 D.(n-1)2解析:因?yàn)閿?shù)列{an}為等比數(shù)列,所以a5·a2n-5=an由a5·a2n-5=22n(n≥3),得an2=22又因?yàn)閍n>0,所以an=2n,所以log2a1+log2a3+…+log2a2n-1=1+3+…+(2n-1)=n2,故選C.答案:C3.已知等比數(shù)列{an}是遞減數(shù)列,前n項(xiàng)的積為Tn,若T13=4T9,則a8a15=().A.±2 B.±4 C.2 D.4解析:∵T13=4T9,∴a1a2…a9a10a11a12a13=4a1a2…a9.∴a10a11a12a13=4.又a10a13=a11a12=a8a15,∴(a8a15)2=4,∴a8a15=±2.∵{an}為遞減數(shù)列,∴q>0.∴a8a15=2.答案:C4.在公差不為零的等差數(shù)列{an}中,2a3-a72+2a11=0,數(shù)列{bn}是等比數(shù)列,且b7=a7,則b6b8=(A.16 B.14 C.4 D.49解析:∵2a3-a72+2a11=2(a3+a11)-a72=4a7-a72=0,又b7=a7≠0,∴b7=a7=4.∴b6b答案:A5.設(shè){an}是公比為q的等比數(shù)列,|q|>1,令bn=an+1(n=1,2,…),若數(shù)列{bn}有連續(xù)四項(xiàng)在集合{-53,-23,19,37,82}中,則6q=.

解析:∵bn=an+1,且數(shù)列{bn}中有連續(xù)四項(xiàng)在集合{-53,-23,19,37,82}中,∴an∈{-54,-24,18,36,81}.∵數(shù)列{an}是公比為q的等比數(shù)列,且|q|>1,∴數(shù)列{an}中的項(xiàng)分別為-24,36,-54,81.∴6q=6×36-24=-答案:-96.在由9個(gè)正數(shù)組成的3行3列方陣a11a12a13a21a22a23a31a32a33中,每行三個(gè)數(shù)成等比數(shù)列,且a11a12a13,a21a22a23,a解析:由題意,知a11a12a13=a123=8,a21a22a23=a223,a31a32a33又a11a12a13,a21a22a23,a31a32a33成等差數(shù)列,所以2a223=a123+a答案:37.在等差數(shù)列{an}中,公差d≠0,a2是a1與a4的等比中項(xiàng).已知數(shù)列a1,a3,ak1,ak2,…,akn,…成等比數(shù)列,求數(shù)列解:依題意得an=a1+(n-1)d,a22=a1a∴(a1+d)2=a1(a1+3d),整理得d2=a1d,∵d≠0,∴d=a1,得an=nd.∴由已知得d,3d,k1d,k2d,…,knd,…是等比數(shù)列.又d≠0,∴數(shù)列1,3,k1,k2,…,kn,…也是等比數(shù)列,首項(xiàng)為1,公比q=31=3,由此得k1=9等比數(shù)列{kn}的首項(xiàng)k1=9,公比q=3,∴kn=9×qn-1=3n+1(n=1,2,3,…),即得到數(shù)列{kn}的通項(xiàng)kn=3n+1.8.已知在等差數(shù)列{an}與等比數(shù)