高考数学一轮复习 第六章 数列 考点规范练30 等比数列及其前n项和 文 新人教B版.doc

考点规范练30等比数列及其前n项和基础巩固1.已知等比数列an满足a1=14,a3a5=4(a4-1),则a2=()a.2b.1c.12d.182.在正项等比数列an中,a2,a48是方程2x2-7x+6=0的两个根,则a1a2a25a48a49的值为()a.212b.93c.93d.353.等差数列an的首项为1,公差不为0.若a2,a3,a6成等比数列,则an前6项的和为()a.-24b.-3c.3d.84.已知an为等比数列,a4+a7=2,a5a6=-8,则a1+a10=()a.7b.5c.-5d.-75.等差数列an的公差为2,若a2,a4,a8成等比数列,则an的前n项和sn=()a.n(n+1)b.n(n-1)c.n(n+1)2d.n(n-1)26.设数列an是首项为a1,公差为-1的等差数列,sn为其前n项和.若s1,s2,s4成等比数列,则a1的值为.7.若等差数列an和等比数列bn满足a1=b1=-1,a4=b4=8,则a2b2=.8.已知等比数列an的各项均为实数,其前n项和为sn.若s3=74,s6=634,则a8=.9.已知an是公差为3的等差数列,数列bn满足b1=1,b2=13,anbn+1+bn+1=nbn.(1)求an的通项公式;(2)求bn的前n项和.10.已知等差数列an的前n项和为sn,且s4=4(a3+1),3a3=5a4,数列bn是等比数列,且b1b2=b3,2b1=a5.(1)求数列an,bn的通项公式;(2)求数列|an|的前n项和tn.11.在数列an中,sn为数列an的前n项和,且sn=1+kan(k0,且k1).(1)求通项公式an;(2)当k=-1时,求a12+a22+an2的值.能力提升12.若a,b是函数f(x)=x2-px+q(p0,q0)的两个不同的零点,且a,b,-2这三个数可适当排序后成等差数列,也可适当排序后成等比数列,则p+q的值等于()a.6b.7c.8d.913.设等比数列an满足a1+a3=10,a2+a4=5,则a1a2an的最大值为.14.设数列an的前n项和为sn.已知s2=4,an+1=2sn+1,nn+.(1)求通项公式an;(2)求数列|an-n-2|的前n项和.高考预测15.(2017四川广元二诊)已知数列an的前n项和为sn,且对任意正整数n都有an=34sn+2成立.若bn=log2an,则b1 008=()a.2 017b.2 016c.2 015d.2 01416.已知数列an满足a1=5,a2=5,an+1=an+6an-1(n2).(1)求证:an+1+2an是等比数列;(2)求数列an的通项公式.参考答案考点规范练30等比数列及其前n项和1.c解析a3a5=4(a4-1),a42=4(a4-1),解得a4=2.又a4=a1q3,且a1=14,q=2.a2=a1q=12.2.b解析a2,a48是方程2x2-7x+6=0的两个根,a2a48=3.又a1a49=a2a48=a252=3,a250,a1a2a25a48a49=a255=93.选b.3.a解析设等差数列的公差为d,则d0,a32=a2a6,即(1+2d)2=(1+d)(1+5d),解得d=-2,所以s6=61+652(-2)=-24,故选a.4.d解析an为等比数列,a5a6=a4a7=-8.联立a4+a7=2,a4a7=-8,可解得a4=4,a7=-2或a4=-2,a7=4,当a4=4,a7=-2时,q3=-12,故a1+a10=a4q3+a7q3=-7;当a4=-2,a7=4时,q3=-2,故a1+a10=a4q3+a7q3=-7.综上可知,a1+a10=-7.5.a解析a2,a4,a8成等比数列,a42=a2a8,即(a1+6)2=(a1+2)(a1+14),解得a1=2.sn=na1+n(n-1)2d=2n+n2-n=n2+n=n(n+1).故选a.6.-12解析由已知得s1=a1,s2=a1+a2=2a1-1,s4=4a1+432(-1)=4a1-6,而s1,s2,s4成等比数列,(2a1-1)2=a1(4a1-6),整理得2a1+1=0,解得a1=-12.7.1解析设等差数列an的公差为d,等比数列bn的公比为q,由题意知-1+3d=-q3=8,即-1+3d=8,-q3=8,解得d=3,q=-2.故a2b2=-1+3-1(-2)=1.8.32解析设该等比数列的公比为q,则s6-s3=634-74=14,即a4+a5+a6=14.s3=74,a1+a2+a3=74.由得(a1+a2+a3)q3=14,q3=1474=8,即q=2.a1+2a1+4a1=74,a1=14,a8=a1q7=1427=32.9.解(1)由已知,a1b2+b2=b1,b1=1,b2=13,得a1=2.所以数列an是首项为2,公差为3的等差数列,通项公式为an=3n-1.(2)由(1)和anbn+1+bn+1=nbn得bn+1=bn3,因此bn是首项为1,公比为13的等比数列.记bn的前n项和为sn,则sn=1-13n1-13=32-123n-1.10.解(1)设等差数列an的公差为d.s4=4(a3+1),3a3=5a4,4a1+6d=4(a1+2d+1),3a1+6d=5a1+15d,解得a1=9,d=-2.an=11-2n.设数列bn的公比为q.b1b2=b3,2b1=a5,b12q=b1q2,2b1=1,解得b1=12,q=12.bn=12n.(2)由(1)知,sn=10n-n2.由an=11-2n0可知n5.5,即a10,a20,a50,a60,a70,an0,q0)的两个不同的零点,a+b=p,ab=q.p0,q0,a0,b0.又a,b,-2这三个数可适当排序后成等差数列,也可适当排序后成等比数列,2b=a-2,ab=4或2a=b-2,ab=4.解得a=4,b=1;解得a=1,b=4.p=a+b=5,q=14=4.p+q=9.故选d.13.64解析由已知a1+a3=10,a2+a4=a1q+a3q=5,两式相除得a1+a3q(a1+a3)=105,解得q=12,a1=8,所以a1a2an=8n121+2+(n-1)=2-12n2+7n2,抛物线f(n)=-12n2+72n的对称轴为n=-722-12=3.5,又nn+,所以当n=3或n=4时,a1a2an取最大值为2-1232+732=26=64.14.解(1)由题意得a1+a2=4,a2=2a1+1,则a1=1,a2=3.又当n2时,由an+1-an=(2sn+1)-(2sn-1+1)=2an,得an+1=3an.所以,数列an的通项公式为an=3n-1,nn+.(2)设bn=|3n-1-n-2|,nn+,b1=2,b2=1.当n3时,由于3n-1n+2,故bn=3n-1-n-2,n3.设数列bn的前n项和为tn,则t1=2,t2=3.当n3时,tn=3+9(1-3n-2)1-3-(n+7)(n-2)2=3n-n2-5n+112,所以tn=2,n=1,3n-n2-5n+112,n2,nn+.15.a解析在an=34sn+2中,令n=1得a1=8,an=34sn+2成立,an+1=34sn+1+2成立,两式相减得an+1-an=34an+1,an+1=4an,又a10,数列an为等比数列,an=84n-1=22n+1,bn=log2an=2n+1,b1008=2017,故选a.16.(1)证明an+1=an+6an-1(n2),an+1+2an=3an+6an-1=3(an+2an-1)(n2).又a1=5,a2=5,a2+2a1=15,an+2an-10(n2),an+1+2anan+2an-1=3(n2),数列a