2019高考数学复习单元质检卷六数列A理新人教B版.docx

单元质检卷六数列(A)(时间:45分钟满分:100分)一、选择题(本大题共6小题,每小题7分,共42分)1.已知等差数列an的前n项和为Sn,a4=15,S5=55,则数列an的公差是()A.B.4C.-4D.-32.公比为的等比数列an的各项都是正数,且a3a11=16,则log2a16=()A.4B.5C.6D.73.(2017宁夏银川二模)在等差数列an中,已知a4=5,a3是a2和a6的等比中项,则数列an的前5项的和为()A.15B.20C.25D.15或254.已知等差数列an和等比数列bn满足:3a1-+3a15=0,且a8=b10,则b3b17=()A.9B.12C.16D.365.(2017湖北武昌1月调研)设公比为q(q0)的等比数列an的前n项和为Sn,若S2=3a2+2,S4=3a4+2,则a1=()A.-2B.-1C.D.6.(2017河南郑州一中质检一,理9)已知数列an满足a1a2a3an=(nN+),且对任意nN+都有+1,nN+,Sn+1+Sn-1=2(Sn+1).(1)求数列an的通项公式;(2)设bn=,求bn的前n项和Tn.10.(15分)数列an满足an=6-(nN+,n2).(1)求证:数列是等差数列;(2)若a1=6,求数列lg an的前999项的和.11.(15分)(2017湖南长郡中学模拟6,理17)已知在数列an中,Sn为其前n项和,若an0,且4Sn=+2an+1(nN+),数列bn为等比数列,公比q1,b1=a1,且2b2,b4,3b3成等差数列.(1)求an与bn的通项公式;(2)令cn=,若cn的前项和为Tn,求证:Tn0,解得q=,代入a1(1+q)=3a1q+2,解得a1=-1.6.D数列an满足a1a2a3an=(nN+),当n=1时,a1=2,当n2时,a1a2a3an-1=,可得an=22n-1.,数列为等比数列,首项为,公比为.+.对任意nN+都有+1,nN+,Sn+1+Sn-1=2(Sn+1),Sn+2+Sn=2(Sn+1+1),相减可得an+2+an=2an+1.(*)又当n=2时,S3+S1=2(S2+1),即2a1+a2+a3=2(a1+a2+1),a1=2,a2=4,解得a3=6.当n=1时(*)也满足.数列an是等差数列,公差为2,an=2+2(n-1)=2n.(2)bn=,bn的前n项和Tn=+,Tn=+,上面两式相减可得Tn=+,Tn=.10.(1)证明 (n2).数列是等差数列.(2)解 a1=6,由(1)知,(n-1)=.an=(nN+),lg an=lg(n+1)-lg n+lg 3(nN+).设数列lg an的前999项的和为S,则S=999lg 3+(lg 2-lg 1+lg 3-lg 2+lg1 000-lg999)=999lg 3+lg1 000=3+999lg 3.11.(1)解 因为4Sn=+2an+1(nN+),所以当n=1时,4a1=+2a1+1,解得a1=1.当n2时,4Sn-1=+2an-1+1,-得4an=(an+1)2-(an-1+1)2,整理得(an+an-1)(an-an-1-2)=0,又an0,an-an-1-2=0,即an-an-1=2,数列an是等差数列,公差为2.an=1+2(n-1)=2n-1.b1=a1=1,2b2,b4,3b3成等差数列,2b4=2b2+3b3.2b2q2=2b2+3b2q,整理