高考数学理北师大版一轮复习测评8-5-1等差与等比数列的综合问题

核心素养测评四十一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(a2a1)= ()A.8 B.8 C.±8 D.QUOTE【解析】选A.由1,a1,a2,9成等差数列,得公差d=a2a1=QUOTE=QUOTE,由1,b1,b2,b3,9成等比数列,得QUOTE=1×9,所以b2=±3,当b2=3时,1,b1,3成等比数列,此时QUOTE=1×(3)无解,所以b2=3,所以b2(a2a1)=3×QUOTE=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.1 B.0 C.2 D.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,q≠0,由a1=b1=1,a5=b3,可得1+4d=q2,则a9=1+8d=1+2(q21)=2q21>1,可得a9能取到的最小整数是0.3.已知在等差数列{an}中,a1>0,d>0,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.S4>T4 B.S4<T4C.S4=T4 D.S4≤T4【解析】选A.设等比数列{bn}的公比为q,则由题意可得q>1,数列{bn}单调递增,又S4T4=a2+a3(b2+b3)=a1+a4a1qQUOTE=a1(1q)+a4QUOTE=QUOTE(a4a1q)=QUOTE(b4b2)>0,所以S4>T4.【一题多解】选A.不妨取an=7n4,则等比数列{bn}的公比q=QUOTE=2,所以S4=54,T4=QUOTE=45,显然S4>T4.4.已知a,b,c成等比数列,a,m,b和b,n,c分别成两个等差数列,则QUOTE+QUOTE等于导学号()A.4 B.3 C.2 D.1【解析】选C.由题意得b2=ac,2m=a+b,2n=b+c,则QUOTE+QUOTE=QUOTE=QUOTE=QUOTE=2.【一题多解】解答本题,还有以下解法:特殊值法:选C.因为a,b,c成等比数列,所以令a=2,b=4,c=8,又a,m,b和b,n,c分别成两个等差数列,则m=QUOTE=3,n=QUOTE=6,因此QUOTE+QUOTE=QUOTE+QUOTE=2.二、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3S23S1=S3S2,则3a2=a3,得公比q=3,所以an=a1qn1=3n1.答案:3n16.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QUOTE=________________.

【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QUOTE=a2a8,所以(a1+3d)2=(a1+d)(a1+7d),所以d2=a1d,因为d≠0,所以d=a1,所以QUOTE=QUOTE=3.答案:37.已知等差数列{an}满足a3=7,a5+a7=26.若bn=QUOTE(n∈N*),则数列{bn}的最大项为____________,最小项为________________.

【解析】因为a5+a7=26,所以a6=13,因为a3=7,所以3d=6,d=2,所以an=a3+QUOTEd=7+2QUOTE=2n+1,所以bn=QUOTE=QUOTE=1+QUOTE,又因为当n=1,2,3时,数列{bn}递减且QUOTE<0,当n≥4时,数列{bn}递减且QUOTE>0,所以数列{bn}的最大项为b4=8,最小项为b3=6.答案:868.已知等差数列QUOTE的公差d≠0,且a1,a3,a13成等比数列,若a1=1,Sn为数列QUOTE的前n项和,则QUOTE的最小值为________________. 导学号

【解析】依题意:因为a1,a3,a13成等比数列,a1=1,所以QUOTE=a1a13,所以(1+2d)2=1+12d,d≠0,解得d=2.可得an=2n1,Sn=n2,则QUOTE=QUOTE=QUOTE=n+2+QUOTE4≥4,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Ⅱ)已知数列{an}和{bn}满足a1=1,b1=0,4an+1=3anbn+4,4bn+1=3bnan4.(1)证明:{an+bn}是等比数列,{anbn}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=QUOTE(an+bn).又因为a1+b1=1,所以QUOTE是首项为1,公比为QUOTE的等比数列.由题设得4(an+1bn+1)=4(anbn)+8,即an+1bn+1=anbn+2.又因为a1b1=1,所以QUOTE是首项为1,公差为2的等差数列.(2)由(1)知,an+bn=QUOTE,anbn=2n1.所以an=QUOTE[(an+bn)+(anbn)]=QUOTE+nQUOTE,bn=QUOTE[(an+bn)(anbn)]=QUOTEn+QUOTE.10.已知等差数列{an}前三项的和为3,前三项的积为8.导学号(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QUOTE解得QUOTE或QUOTE所以由等差数列通项公式可得an=23(n1)=3n+5或an=4+3(n1)=3n7.故an=3n+5或an=3n7.(2)当an=3n+5时,a2,a3,a1分别为1,4,2,不成等比数列;当an=3n7时,a2,a3,a1分别为1,2,4,成等比数列,满足条件.故|an|=|3n7|=QUOTE记数列{|an|}的前n项和为Sn.当n=1时,S1=|a1|=4;