等比數(shù)列 同步練習(xí) 含解析

等比數(shù)列同步練習(xí)一、等比數(shù)列中基本量的運(yùn)算1.已知{an}是等比數(shù)列,a2=2,a5=,則公比q等于

(

)A.-

B.-2

C.2

D.答案:D解析:=q3=,∴q=.2.已知等比數(shù)列{an}中,a1=32,公比q=-,則a6等于(

)A.1

B.-1

C.2

D.答案:B解析:由題知a6=a1q5=32×=-1,故選B.3.(2015福建寧德五校聯(lián)考,7)已知等比數(shù)列{an}中,=2,a4=8,則a6=(

)A.31

B.32

C.63

D.64答案:B解析:設(shè)等比數(shù)列{an}的公比為q,由=2,a4=8,得解得所以a6=a1q5=25=32.故選B.4.(2015山東濰坊四縣聯(lián)考,3)已知等差數(shù)列{an}的公差為2,若a1,a3,a4成等比數(shù)列,則a2等于(

)A.-4

B.-6

C.-8

D.-10答案:B解析:∵等差數(shù)列{an}的公差為2,a1,a3,a4成等比數(shù)列,∴(a1+4)2=a1(a1+6),∴a1=-8,∴a2=-6.故選B.5.(2015江西吉安聯(lián)考,2)已知等比數(shù)列{an}的公比q=-,則等于(

)A.-3

B.-

C.3

D.答案:A解析:∵等比數(shù)列{an}的公比q=-,∴=-3.故選A.二、等比中項(xiàng)及應(yīng)用6.2+和2-的等比中項(xiàng)是

.

答案:±1解析:設(shè)A為等比中項(xiàng),則A2=(2+)(2-)=1,∴A=±1.7.已知等比數(shù)列{an}的各項(xiàng)均為正數(shù),它的前三項(xiàng)依次為1,a+1,2a+5,則數(shù)列{an}的通項(xiàng)公式an=

.

答案:3n-1解析:由題意,知(a+1)2=2a+5,∴a2=4.∵{an}的各項(xiàng)均為正數(shù),∴a+1>0且2a+5>0.∴a=2.∴a+1=3.∴q==3.∴an=3n-1.三、等比數(shù)列的判定8.給出下列數(shù)列:①2,2,4,8,16,32,…;②在數(shù)列{an}中,=2,=2;③常數(shù)列c,c,c,c,….其中等比數(shù)列的個(gè)數(shù)為

.

答案:0解析:①不是等比數(shù)列,因?yàn)?②不一定是等比數(shù)列,因?yàn)椴恢赖闹?即使=2,數(shù)列{an}也未必是等比數(shù)列;③不一定是等比數(shù)列,當(dāng)c=0時(shí),數(shù)列不是等比數(shù)列.故填0.9.設(shè){an}是公比為q的等比數(shù)列,設(shè)q≠1,證明數(shù)列{an+1}不是等比數(shù)列.證明:假設(shè){an+1}是等比數(shù)列,則對(duì)任意的k∈N*,(ak+1+1)2=(ak+1)(ak+2+1),+2ak+1+1=akak+2+ak+ak+2+1,q2k+2a1qk=a1qk-1·a1qk+1+a1qk-1+a1qk+1,因?yàn)閍1≠0,所以2qk=qk-1+qk+1.因?yàn)閝≠0,所以q2-2q+1=0,解得q=1,這與已知矛盾.所以假設(shè)不成立,故{an+1}不是等比數(shù)列.(建議用時(shí):30分鐘)1.已知在等比數(shù)列{an}中,a1+a3=10,a4+a6=,則該等比數(shù)列的公比為(

)A.

B.

C.2

D.8答案:B解析:因?yàn)?a1+a3)q3=a4+a6,所以q3=,即q=,選B.2.若等比數(shù)列的首項(xiàng)為,末項(xiàng)為,公比為,則這個(gè)數(shù)列的項(xiàng)數(shù)為(

)A.3

B.4

C.5

D.6答案:B解析:∵a1=,an=,q=,∴,則n=4.3.已知等比數(shù)列{an}中,a1=3,8=an+1·an+2,則a3=

(

)A.48

B.12

C.6

D.2答案:B解析:設(shè)數(shù)列{an}的公比為q,則由8=an+1an+2,得8=a2a3,即8q3,∴q=2.∴a3=a1q2=3×4=12.4.如果-1,a,b,c,-9成等比數(shù)列,那么(

)A.b=3,ac=9

B.b=-3,ac=9C.b=3,ac=-9

D.b=-3,ac=-9答案:B解析:∵-1,a,b,c,-9成等比數(shù)列,∴b2=(-1)×(-9)=9.又∵a2=-1×b=-b,∴b=-3.又b2=ac,∴a與c同號(hào).∴ac=9.5.已知1既是a2與b2的等比中項(xiàng),又是的等差中項(xiàng),則的值是(

)A.1或

B.1或-C.1或

D.1或-答案:D解析:由題意得,a2b2=(ab)2=1,=2,∴又,∴其值為1或-.6.設(shè)a1=2,數(shù)列{1+2an}是公比為2的等比數(shù)列,則a6等于

.

答案:79.5解析:∵1+2an=(1+2a1)×2n-1,∴1+2a6=5×25,∴a6==79.5.7.已知等差數(shù)列{an}的公差d≠0,它的第1,5,17項(xiàng)順次成等比數(shù)列,則這個(gè)等比數(shù)列的公比是

.

答案:3解析:由已知=a1·a17,∴(a1+4d)2=a1(a1+16d).∴a1=2d.∴公比q==3.8.某林場(chǎng)的樹木每年以25%的增長(zhǎng)率增長(zhǎng),則第10年末的樹木總量是今年的

倍.

答案:1.259解析:設(shè)這個(gè)林場(chǎng)今年的樹木總量是m,第n年末的樹木總量為an,則an+1=an+an×25%=1.25an.則=1.25.則數(shù)列{an}是公比q=1.25的等比數(shù)列.則a10=a1q9=1.259m.所以=1.259.9.等比數(shù)列的前三項(xiàng)和為168,a2-a5=42,求a5,a7的等比中項(xiàng).解:由題意知②÷①得q(1-q)=,∴q=.∴a1==96.又∵a6=a1q5,∴a6=96×=3,∴a5,a7的等比中項(xiàng)a6=3.10.