3.1等比數(shù)列的概念及其通項(xiàng)公式第1課時(shí)等比數(shù)列的概念及其通項(xiàng)公式課后訓(xùn)練鞏固提升1.已知{an}是等比數(shù)列,a2=2,a5=14,則公比q=()A.-12 B.-2 C.2 D.解析:∵q3=a5a2=1答案:D2.等比數(shù)列x,3x+3,6x+6,…的第四項(xiàng)等于().A.-24 B.0 C.12 D.24解析:由題可得(3x+3)2=x(6x+6),解得x=-3或x=-1(舍),∴等比數(shù)列的前三項(xiàng)為-3,-6,-12.∴第四項(xiàng)為-24,故選A.答案:A3.在等比數(shù)列{an}中,a1=1,a10=3,則a2a3a4a5a6a7a8a9等于().A.81 B.27327 C.3 D.解析:a10=a1q9=3,又a1=1,∴q9=3.而a2a3…a9=a18q1+2+…+8=q36=(q9)4=答案:A4.已知由正數(shù)組成的等比數(shù)列{an}中,公比q=2,a1·a2·a3·…·a30=245,則a1·a4·a7·…·a28的值為().A.25 B.210 C.215 D.220解析:∵公比q=2,∴a2·a5·a8·…·a29=a1·a4·a7·…·a28·210,a3·a6·a9·…·a30=a1·a4·a7·…·a28·220,∴a1·a4·a7·…·a28=25.答案:A5.(多選題)設(shè)數(shù)列{an}為等比數(shù)列,則下面四個(gè)選項(xiàng)中的數(shù)列是等比數(shù)列的是().A.{an3} B.{pan}(pC.{an·an+1} D.{an+an+1}解析:A項(xiàng)中,因?yàn)閧an}為等比數(shù)列,所以an+1an=q,所以an+13a因?yàn)閍13≠0,所以{anB項(xiàng)中,pan+1pan=an+1an=q,C項(xiàng)中,an+2·an+1an+1·an=an+2an=q2,且D項(xiàng)中,1,-1,1,-1,…是等比數(shù)列,但an+an+1=0,所以{an+an+1}不一定是等比數(shù)列.故選ABC.答案:ABC6.已知等比數(shù)列{an}中,a3=6,a10=768,則該數(shù)列的通項(xiàng)an=.

解析:記公比為q.由已知得a1q2=6,a1q9=768,∴q7=128=27,故q=2,a1=32,∴an=a1·qn-1=3·2n-2答案:3·2n-27.設(shè)數(shù)列{an}是首項(xiàng)為1,公比為-2的等比數(shù)列,則a1+|a2|+a3+|a4|=.

解析:由題可知,a1+|a2|+a3+|a4|=1+|1×(-2)|+1×22+|1×(-2)3|=1+2+4+8=15.答案:158.已知-7,a1,a2,-1四個(gè)實(shí)數(shù)成等差數(shù)列,-4,b1,b2,b3,-1五個(gè)實(shí)數(shù)成等比數(shù)列,則b2=;a2-a1解析:由題意得a2-a1=13[(-1)-(-7)]=設(shè)等比數(shù)列的公比為q,則(-4)q2=b2,b2q2=-1,∴b22=(-4)×(-1)=4,且b2<0,∴b2=-∴a2-a答案:2-19.在等比數(shù)列{an}中,a3=32,a5=8,(1)求數(shù)列{an}的通項(xiàng)公式an;(2)若an=12,求n解:(1)因?yàn)閍5=a3q2,所以q2=a5a3=14當(dāng)q=12時(shí),an=a3qn-3=32×12n-3當(dāng)q=-12時(shí),an=a3qn-3=32×-所以an=28-n或an=32×-1(2)當(dāng)an=12時(shí),28-n=12或32×解得n=9.10.已知數(shù)列{an},{bn}滿足下列條件:a1=0,a2=1,an+2=an+an+12,bn(1)求證:數(shù)列{bn}是等比數(shù)列;(2)求數(shù)列{bn}的通項(xiàng)公式.(1)證明:∵