等比數(shù)列1.已知等比數(shù)列{an},a2a9=8,a5=2,則公比q為()A.12 B.2 C.14 D2.在等比數(shù)列{an}中,a4=24,a6=6,則a5=()A.12 B.-12 C.±12 D.153.在等比數(shù)列{an}中,若a2a4a6a8=16,則a5=()A.-2 B.3 C.-2或2 D.44.對(duì)任意等比數(shù)列{an},下列說(shuō)法一定正確的是()A.a1,a3,a9成等比數(shù)列B.a2,a3,a6成等比數(shù)列C.a2,a4,a8成等比數(shù)列D.a3,a6,a9成等比數(shù)列5.(多選題)已知在等比數(shù)列{an}中,a1=1,q=2,則()A.數(shù)列{a2n}是等比數(shù)列B.數(shù)列{1aC.數(shù)列{log2an}是等差數(shù)列D.數(shù)列{an}是遞增數(shù)列6.已知{an}是等比數(shù)列,a1=12,a2=4,則a3=,a1a2a3a4a5a6=.B級(jí)關(guān)鍵能力提升練7.設(shè)無(wú)窮等比數(shù)列{an},則“0<a2<a1”是“{an}為遞減數(shù)列”的()A.充分而不必要條件B.必要而不充分條件C.充要條件D.既不充分又不必要條件8.在正項(xiàng)等比數(shù)列{an}中,若a3a7a8=8,a2+a10=5,則公比q=()A.212 B.2C.214 D.29.兩個(gè)公比均不為1的等比數(shù)列{an},{bn},其前n項(xiàng)的乘積分別為An,Bn.若a5b5=2,則A9A.512 B.32 C.8 D.210.(多選題)設(shè){an}是公比為2的等比數(shù)列,則下列四個(gè)選項(xiàng)正確的有()A.{1an}是公比為B.{a2n}是公比為4的等比數(shù)列C.{2an}是公比為4的等比數(shù)列D.{anan+1}是公比為2的等比數(shù)列11.記Sn為數(shù)列{an}的前n項(xiàng)和,且滿(mǎn)足a1<0,Sn=λan-1,若數(shù)列{an}為遞增數(shù)列,則實(shí)數(shù)λ的取值范圍為()A.(1,+∞) B.(-∞,0)C.(0,1) D.(-∞,0)∪(1,+∞)12.已知在各項(xiàng)都為正數(shù)的等比數(shù)列{an}中,a2a4=4,a1+a2+a3=14,則滿(mǎn)足anan+1an+2>19的最大正整數(shù)n的值為.13.在等比數(shù)列{an}中,若a7+a8+a9+a10=158,a8a9=-98,則1a714.已知等比數(shù)列{an}為遞增數(shù)列,且a52=a10,2(an+an-2)=5an-1,求數(shù)列{anC級(jí)學(xué)科素養(yǎng)創(chuàng)新練15.在各項(xiàng)都為正數(shù)的等比數(shù)列{an}中,已知a1=512,其前n項(xiàng)積為T(mén)n,且T13=T6,則Tn取得最大值時(shí),n的值是()A.9 B.8或9C.10或11 D.9或10等比數(shù)列1.B因?yàn)閍2a9=8,所以a5a6=a2a9=8.又因?yàn)閍5=2,所以a6=4,所以公比q=2,故選B.2.C根據(jù)題意,可知a52=a4a6=24×6=144,解得a5=3.C由等比數(shù)列的性質(zhì),知a2a4a6a8=a54=16,可得a5=4.D設(shè)等比數(shù)列{an}的公比為q,則a3=a1q2,a6=a1q5,a9=a1q8,滿(mǎn)足(a1q5)2=a1q2·a1q8,即(a6)2=a3a9.故D正確.5.ACD在等比數(shù)列{an}中,a1=1,q=2,所以an=2n-1.a2n=22n-1,數(shù)列{a2n}依舊是等比數(shù)列,選項(xiàng)A正確;1an=12n-1=log2an=log22n-1=n-1,顯然數(shù)列{log2an}是等差數(shù)列,選項(xiàng)C正確;由于a1>0,q>1,因此選項(xiàng)D正確.6.32239因?yàn)閿?shù)列{an}是等比數(shù)列,且a1=12,a2=所以等比數(shù)列{an}的公比q=a2a1=8,所以a3=a2q=4×8=32.所以a1a2a3a4a5a6=(a3)5·a1q5=325×17.A{an}為無(wú)窮等比數(shù)列,若0<a2<a1,則公比q滿(mǎn)足0<q=a2a1<1,所以{an}為遞減數(shù)列;反之,若無(wú)窮等比數(shù)列{如-18,-14,-12,-1,-2,…,所以不一定得到0<a2故“0<a2<a1”是“{an}為遞減數(shù)列”的充分而不必要條件,故選A.8.D由a3a7a8=8得a1q2·a1q6·a1q7=(a1q5)3=a63=8,即a6=2.則a2a10=(a6)2=4,又a∵q>0,∴q=(a10a2)

18=4189.A因?yàn)锳9=a1a2a3…a9=a59,B9=b1b2b3…b9=所以A9B9=(a5b5)910.AB由于數(shù)列{an}是公比為2的等比數(shù)列,則對(duì)任意的n∈N+,an≠0,且公比為q=an+1a1an+11aa2n+2a2n=q22an+12aan+1an+2anan+1=an11.B由Sn=λan-1,及Sn-1=λan-1-1(n≥2),作差可得an=λan-λan-1(n≥2),即(λ-1)an=λan-1(n≥2).因?yàn)閍1<0,所以λ≠1,所以anan-1=λ若數(shù)列{an}為遞增數(shù)列,則0<λλ-1<1,解得λ<012.4∵a2a4=4=a32,且a3>0,∴a3=又a1+a2+a3=2q2+2∴1q=-3(舍去)或1q=2,即q=12,a1又an=a1qn-1=8×(12)n-1=(12)n-∴anan+1an+2=(12)3n-9>19,即23n-9∴n的最大值為4.13.-53因?yàn)?又a8a9=a7a10,所以1a7+14.解設(shè)數(shù)列{an}的公比為q,首項(xiàng)為a1.因?yàn)閍52=a10,2(an+an-2)=5an-所以a由①得a1=q,由②得q=2或q=12又因?yàn)閿?shù)列{an}為遞增數(shù)列,所以a1=q=2,所以an=2n.15.D(方法1)∵a1=512,T13=T6,∴a7a8a9a10a11a12a13=1,∴a107=1,∴a10=a