（新课改地区）2021版高考数学一轮复习 核心素养测评三十五 等差与等比数列的综合问题 新人教B版

核心素养测评三十五 等差与等比数列的综合问题(30分钟60分)一、选择题(每小题5分,共20分)1.已知1,a1,a2,9四个实数成等差数列,1,b1,b2,b3,9五个数成等比数列,则b2(a2-a1)=()a.8 b.-8 c.8d.98【解析】选a.由1,a1,a2,9成等差数列,得公差d=a2-a1=9-14-1=83,由1,b1,b2,b3,9成等比数列,得b22=19,所以b2=3,当b2=-3时,1,b1,-3成等比数列,此时b12=1(-3)无解,所以b2=3,所以b2(a2-a1)=383=8.2.等差数列an,等比数列bn,满足a1=b1=1,a5=b3,则a9能取到的最小整数是()a.-1b.0c.2d.3【解析】选b.等差数列an的公差设为d,等比数列bn的公比设为q,q0,由a1=b1=1,a5=b3,可得1+4d=q2,则a9=1+8d=1+2(q2-1)=2q2-1-1,可得a9能取到的最小整数是0.3.已知在等差数列an中,a10,d0,前n项和为sn,等比数列bn满足b1=a1,b4=a4,前n项和为tn,则()a.s4t4b.s41,数列bn单调递增,又s4-t4=a2+a3-(b2+b3)=a1+a4-a1q-a4q=a1(1-q)+a41-1q=q-1q(a4-a1q)=q-1q(b4-b2)0,所以s4t4.【一题多解】选a.不妨取an=7n-4,则等比数列bn的公比q=3a4a1=2,所以s4=54,t4=b1(1-q4)1-q=45,显然s4t4.4.(多选)等比数列an的前n项和为sn,若对任意的正整数n,sn+2=4sn+3恒成立,则a1的值为()a.3b.1c.-3d.-1【解析】选bc.设等比数列an的公比为q,当q=1时,sn+2=(n+2)a1,sn=na1,由sn+2=4sn+3得,(n+2)a1=4na1+3,即3a1n=2a1-3,若对任意的正整数n,3a1n=2a1-3恒成立,则a1=0且2a1-3=0,矛盾,所以q1,所以sn=a1(1-qn)1-q,sn+2=a1(1-qn+2)1-q,代入sn+2=4sn+3并化简得a1(4-q2)qn=3+3a1-3q,若对任意的正整数n该等式恒成立,则有4-q2=0,3+3a1-3q=0,解得a1=1,q=2或a1=-3,q=-2,故a1=1或-3.二、填空题(每小题5分,共20分)5.sn为等比数列an的前n项和.若a1=1,且3s1 ,2s2,s3成等差数列,则an=_.【解析】由3s1,2s2,s3成等差数列,得4s2=3s1+s3,即3s2-3s1=s3-s2,则3a2=a3,得公比q=3,所以an=a1qn-1=3n-1.答案:3n-16.已知等差数列an的公差和首项都不等于0,且a2,a4,a8成等比数列,则a1+a5+a9a2+a3=_.【解析】设公差为d,因为在等差数列an中,a2, a4,a8成等比数列,所以a42=a2a8,所以(a1+3d)2=(a1+d)(a1+7d),所以d2=a1d,因为d0,所以d=a1,所以a1+a5+a9a2+a3=15a15a1=3.答案:37.(2020银川模拟)已知an是等差数列,a1=1,公差d0,sn为其前n项和,若a1,a2,a5成等比数列,则s8=_.【解析】因为a1,a2,a5成等比数列,则a22=a1a5,即(1+d)2=1(1+4d),解得d=2.所以an=1+(n-1)2=2n-1,a8=28-1=15,s8=(a1+a8)82=4(1+15)=64.答案:648.已知等差数列an的公差d0,且a1,a3,a13成等比数列,若a1=1,sn为数列an的前n项和,则2sn+24an+5的最小值为_.【解析】依题意:因为a1,a3,a13成等比数列,a1=1,所以a32=a1a13,所以(1+2d)2=1+12d,d0,解得d=2.可得an=2n-1,sn=n2,则2sn+24an+5=n2+12n+2=(n+2)2-4n+2+16n+2=n+2+16n+2-44,当且仅当n=2时,等号成立.答案:4三、解答题(每小题10分,共20分)9.(2019全国卷)已知数列an和bn满足a1=1,b1=0,4an+1=3an-bn+4,4bn+1=3bn-an-4.(1)证明:an+bn是等比数列,an-bn是等差数列.(2)求an和bn的通项公式.【解析】(1)由题设得4(an+1+bn+1)=2(an+bn),即an+1+bn+1=12(an+bn).又因为a1+b1=1,所以an+bn是首项为1,公比为12的等比数列.由题设得4(an+1-bn+1)=4(an-bn)+8,即an+1-bn+1=an-bn+2.又因为a1-b1=1,所以an-bn是首项为1,公差为2的等差数列.(2)由(1)知,an+bn=12n-1,an-bn=2n-1.所以an=12(an+bn)+(an-bn)=12n+n-12,bn=12(an+bn)-(an-bn)=12n-n+12.10.已知等差数列an前三项的和为-3,前三项的积为8.(1)求数列an的通项公式.(2)若a2,a3,a1成等比数列,求数列|an|的前n项和sn.【解析】(1)设等差数列an的公差为d,则a2=a1+d,a3=a1+2d.由题意得3a1+3d=-3,a1(a1+d)(a1+2d)=8,解得a1=2,d=-3,或a1=-4,d=3.所以由等差数列通项公式可得an=2-3(n-1)=-3n+5或an=-4+3(n-1)=3n-7.故an=-3n+5或an=3n-7.(2)当an=-3n+5时,a2,a3,a1分别为-1,-4,2,不成等比数列;当an=3n-7时,a2,a3,a1分别为-1,2,-4,成等比数列,满足条件.故|an|=|3n-7|=-3n+7,n=1,2,3n-7,n3.记数列|an|的前n项和为sn.当n=1时,s1=|a1|=4;当n=2时,s2=|a1|+|a2|=5;