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-1-第2课时等比数列的性质及应用1.公比为2的等比数列{an}的各项都是正数,且a3a11=16,则a5等于()A.1B.2C.4D.8解析:∵a3a11==16,且an0,∴a7=4.又a7=a5·q2=4a5,∴a5=1.答案:A2.已知等比数列{an}中,有a3a11=4a7,数列{bn}是等差数列,且b7=a7,则b5+b9等于()A.2B.4C.8D.16解析:等比数列{an}中,a3a11==4a7,解得a7=4,等差数列{bn}中,b5+b9=2b7=2a7=8.答案:C3.若数列{an}是等比数列,则下列数列一定是等比数列的是()A.{lgan}B.{1+an}C.D.{}解析:当an=-1时,lgan与无意义,1+an=0,则选项A,B,D都不符合题意;选项C中,设an=a1qn-1(q是公比),则bn=,则有=常数,即是等比数列.答案:C4.已知各项均为正数的等比数列{an}中,a1a2a3=5,a7a8a9=10,则a4a5a6等于()A.5B.7C.6D.4解析:∵a1a2a3==5,∴a2=.∵a7a8a9==10,∴a8=.∴=a2a8==5.-2-又∵数列{an}各项均为正数,∴a5=5.∴a4a5a6==5=5.答案:A5.等比数列{an}的各项都为正数,且a5a6+a4a7=18,则log3a1+log3a2+…+log3a10等于()A.12B.10C.8D.2+log35解析:a5a6+a4a7=2a5a6=18,所以a5a6=9.所以log3a1+log3a2+…+log3a10=log3(a1a2…a10)=log3[(a1a10)(a2a9)…(a5a6)]=log3[(a5a6)5]=log395=10.答案:B6.已知等比数列{an}为递增数列,若a10,且2(an+an+2)=5an+1,则数列{an}的公比q=.解析:∵数列{an}是等比数列,且2(an+an+2)=5an+1,∴2(an+anq2)=5anq,即2(1+q2)=5q.解方程得q=或q=2.∵a10,数列{an}为递增数列,∴q=2.答案:27.公差不为零的等差数列{an}中,2a3-+2a11=0,数列{bn}是等比数列,且b7=a7,则b6b8=.解析:∵2a3-+2a11=2(a3+a11)-=4a7-=0,又∵b7=a7≠0,∴b7=a7=4.∴b6b8==16.答案:168.在两数1,16之间插入3个数,使它们成等比数列,则中间的数等于.解析:设插入的三个数分别为a,b,c,则b2=16,∴b=±4.设其公比为q,∵b=1·q20,∴b=4.答案:49.已知数列{an}是等比数列,a3+a7=20,a1a9=64,求a11的值.解:∵{an}为等比数列,∴a1·a9=a3·a7=64.又a3+a7=20,∴a3,a7是方程t2-20t+64=0的两个根.解方程,得t1=4,t2=16,∴a3=4,a7=16或a3=16,a7=4.-3-当a3=4时,a3+a7=a3+a3q4=20,∴1+q4=5.∴q4=4.∴a11=a3q8=4×42=64.当a3=16时,a3+a7=a3(1+q4)=20,∴1+q4=.∴q4=.∴a11=a3q8=16×=1.综上可知,a11的值为64或1.10.三个互不相等的数成等差数列,如果适当排列这三个数,又可成为等比数列,且这三个数的和为6,求这三个数.解:由已知,可设这三个数为a-d,a,a+d,则a-d+a+a+d=6,∴a=2,这三个数可表示为2-d,2,2+d,①若2-d为等比中项,则有(2-d)2=2(2+d),解之得d=6或d=0(舍去).此时三个数为-4,2,8.②若2+d为等比中项,则有(2+d)2=2(2-d),解之得d=-6或d=0(舍去).此时三个数为8,2,-4.③若2为等比中项,则22=(2+d)·(2-d),解得d=0(舍去).综上所述,这三个数为-4,2,8或8,2,-4.