1、核心素养测评三十五 等差与等比数列的综合问题一、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.【解析】选A.由1,a1,a2,9成等差数列,得公差d=a2-a1=,由1,b1,b2,b3,9成等比数列,得=19,所以b2=3,当b2=-3时,1,b1,-3成等比数列,此时=1(-3)无解,所以b2=3,所以b2(a2-a1)=3=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
2、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-=a1(1-q)+a4=(a4-a1q)=(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=2,所以S4=54,T4=45,显然S4T4.4.(多
3、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=,Sn+2=,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,则有解得或故a1=1或-3.二、填空题(每小题5分,共20分)5.Sn为等比数列
4、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=_.【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=a2a8,所以(a1+3d)2=(a1+d)(a1+7d),所以d2=a1d,因为d0,所以d=a1,所以=3.答案:37.(2020镇江模拟)已知an是等差数列,a1=1,公差d0,Sn为
5、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=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=4(1+15)=64.答案:648.已知等差数列的公差d0,且a1,a3,a13成等比数列,若a1=1,Sn为数列的前n项和,则的最小值为_.【解析】依题意:因为a1,a3,a13成等比数列,a1=1,所以=a1a13,所以(1+2d)2=1+12d,d0,解得d=2.可得an=2n-1,Sn=n2,则=n+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
6、,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=(an+bn).又因为a1+b1=1,所以是首项为1,公比为的等比数列.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,所以是首项为1,公差为2的等差数列.(2)由(1)知,an+bn=,an-bn=2n-1.所以an=(
7、an+bn)+(an-bn)=+n-,bn=(an+bn)-(an-bn)=-n+.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.由题意得解得或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|=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;当n3时,Sn=S2+|a3|+|a4|+|an|=5+(33-7)+(34-7)+(3n-7)=5+=n2-n+10.当n=2时,满足此式,当n=1时,不满足此式.综上,Sn=
