1.已知等比数列{an}中,a1=32,公比q=- ,则a6等于( )
A.1 B.-1 C.2 D.
解析:由题知a6=a1q5=32× =-1,故选B.
答案:B
2.在等比数列{an}中,a1=1,公比|q|≠1.若am=a1a2a3a4a5,则m等于( )
A.9 B.10 C.11 D.12
解析:∵am=a1a2a3a4a5=q·q2·q3·q4=q10=1×q10,
∴m=11.
答案:C
3.设等差数列{an}的公差d不为0,a1=9d.若ak是a1与a2k的等比中项,则k等于( )
A.2 B.4 C.6 D.8
解析:由题意得an=(n+8)d, =a1a2k,
∴(k+8)2d2=9d(2k+8)d,
解得k=4(k=-2不合题意,舍去).
答案:B
4.若等比数列{an}满足anan+1=16n,则公比q为( )
A.2 B.4 C.8 D.16
解析:∵anan+1=16n,
∴a1a2=16,a2a3=162.
两式相除得 =16,即q2=16.∴q=±4.
∵anan+1=16n>0,∴an,an+1同号,即q>0,
