（福建專用）高考數(shù)學(xué)一輪復(fù)習(xí) 課時(shí)規(guī)范練29 等差數(shù)列及其前n項(xiàng)和 理 新人教A-新人教A高三數(shù)學(xué)試題

課時(shí)規(guī)范練29等差數(shù)列及其前n項(xiàng)和一、基礎(chǔ)鞏固組1.已知等差數(shù)列{an}中,a4+a5=a3,a7=-2,則a9=()A.-8 B.-6 C.-4 D.-2.(2017陜西咸陽二模,理4)《張丘建算經(jīng)》卷上一題為“今有女善織,日益功疾,且從第二天起,每天比前一天多織相同量的布,現(xiàn)在一月(按30天計(jì))共織布390尺,最后一天織布21尺”,則該女第一天織布多少尺?()A.3 B.4 C.5 D.63.已知在每項(xiàng)均大于零的數(shù)列{an}中,首項(xiàng)a1=1,且前n項(xiàng)和Sn滿足SnSn-1-Sn-1Sn=2SnSn-1(n∈NA.638 B.639 C.640 D.6414.已知數(shù)列{an}是等差數(shù)列,a1+a3+a5=105,a2+a4+a6=99,{an}的前n項(xiàng)和為Sn,則使得Sn取最大值時(shí),n的值是()A.18 B.19 C.20 D.215.(2017遼寧沈陽質(zhì)量檢測(cè))設(shè)Sn為等差數(shù)列{an}的前n項(xiàng)和,若a1=1,公差d=2,Sn+2-Sn=36,則n=()A.5 B.6 C.7 D.86.(2017北京豐臺(tái)一模,理10)已知{an}為等差數(shù)列,Sn為其前n項(xiàng)和.若a2=2,S9=9,則a8=.

7.已知在數(shù)列{an}中,a1=1,a2=2,當(dāng)整數(shù)n≥2時(shí),Sn+1+Sn-1=2(Sn+S1)都成立,則S15=.

8.一個(gè)等差數(shù)列的前12項(xiàng)的和為354,前12項(xiàng)中偶數(shù)項(xiàng)的和與奇數(shù)項(xiàng)的和的比為32∶27,則該數(shù)列的公差d=.

9.若數(shù)列{an}的前n項(xiàng)和為Sn,且滿足an+2SnSn-1=0(n≥2),a1=12(1)求證:1S(2)求數(shù)列{an}的通項(xiàng)公式.?導(dǎo)學(xué)號(hào)21500542?10.Sn為等差數(shù)列{an}的前n項(xiàng)和,且a1=1,S7=28.記bn=[lgan],其中[x]表示不超過x的最大整數(shù),如[0.9]=0,[lg99]=1.(1)求b1,b11,b101;(2)求數(shù)列{bn}的前1000項(xiàng)和.二、綜合提升組11.若數(shù)列{an}滿足:a1=19,an+1=an-3(n∈N*),則數(shù)列{an}的前n項(xiàng)和數(shù)值最大時(shí),n的值為()A.6 B.7 C.8 D.912.(2017四川廣元二診,理10)設(shè)等差數(shù)列{an}的前n項(xiàng)和為Sn,若Sm-1=-2,Sm=0,Sm+1=3,其中m≥2,則nSn的最小值為()A.-3 B.-5 C.-6 D.-913.數(shù)列{an}是等差數(shù)列,數(shù)列{bn}滿足bn=anan+1an+2(n∈N*),設(shè)Sn為{bn}的前n項(xiàng)和.若a12=38a5>0,則當(dāng)Sn取得最大值時(shí),n的值等于14.已知公差大于零的等差數(shù)列{an}的前n項(xiàng)和為Sn,且滿足a3·a4=117,a2+a5=22.(1)求通項(xiàng)公式an;(2)求Sn的最小值;(3)若數(shù)列{bn}是等差數(shù)列,且bn=Snn?導(dǎo)學(xué)號(hào)21500543?三、創(chuàng)新應(yīng)用組15.有兩個(gè)等差數(shù)列{an},{bn},其前n項(xiàng)和分別為Sn和Tn,若SnTn=3A.2719 B.1813 C.107課時(shí)規(guī)范練29等差數(shù)列及其前n項(xiàng)和1.B解法一:由已知可得2解得a1=10,d=-2,所以a9=10+(-2)×8=-6.解法二:因?yàn)閍4+a5=a3,所以a3+a6=a3,a6=0,又a7=-2,所以d=-2,a9=-2+(-2)×2=-6.2.C設(shè)第n天織布an尺,則數(shù)列{an}是等差數(shù)列,且S30=390,a30=21,∴S30=302(a1+a30),即390=15(a1+21),解得a1=5.故選C3.C由已知SnSn-1-Sn-1Sn=2S∴{Sn故Sn=2n-1,Sn=(2n-1)2∴a81=S81-S80=1612-1592=640,故選C.4.Ca1+a3+a5=105?a3=35,a2+a4+a6=99?a4=33,則{an}的公差d=33-35=-2,a1=a3-2d=39,Sn=-n2+40n,因此當(dāng)Sn取得最大值時(shí),n=20.5.D解法一:由題知Sn=na1+n(n-1)2d=n+n(n-1)=n2,Sn+2=(n+2)2,由Sn+2-Sn=36,得(n+2)2-n2=4n+解法二:Sn+2-Sn=an+1+an+2=2a1+(2n+1)d=2+2(2n+1)=36,解得n=86.0∵{an}為等差數(shù)列,Sn為其前n項(xiàng)和.a2=2,S9=9,∴解得d=-13,a1=7∴a8=a1+7d=0.7.211由Sn+1+Sn-1=2(Sn+S1)得(Sn+1-Sn)-(Sn-Sn-1)=2S1=2,即an+1-an=2(n≥2),∴數(shù)列{an}從第二項(xiàng)起構(gòu)成以2為首項(xiàng),2為公差的等差數(shù)列,則S15=1+2×14+14×132×8.5設(shè)該等差數(shù)列的前12項(xiàng)中奇數(shù)項(xiàng)的和為S奇,偶數(shù)項(xiàng)的和為S偶,等差數(shù)列的公差為d.由題意得S解得S又S偶-S奇=6d,所以d=192-16269.(1)證明當(dāng)n≥2時(shí),由an+2SnSn-1=0,得Sn-Sn-1=-2SnSn-1,所以1Sn-又1S1=1a(2)解由(1)可得1Sn=2n,∴Sn當(dāng)n≥2時(shí),an=Sn-Sn-1=12n當(dāng)n=1時(shí),a1=12不適合上式故an=110.解(1)設(shè){an}的公差為d,據(jù)已知有7+21d=28,解得d=1.所以{an}的通項(xiàng)公式為an=n.b1=[lg1]=0,b11=[lg11]=1,b101=[lg101]=2.(2)因?yàn)閎n=0所以數(shù)列{bn}的前1000項(xiàng)和為1×90+2×900+3×1=1893.11.B∵a1=19,an+1-an=-3,∴數(shù)列{an}是以19為首項(xiàng),-3為公差的等差數(shù)列.∴an=19+(n-1)×(-3)=22-3n.設(shè){an}的前k項(xiàng)和數(shù)值最大,則有ak≥0,ak∴∴193∵k∈N*,∴k=7.∴滿足條件的n的值為7.12.D由Sm-1=-2,Sm=0,Sm+1=3,得am=2,am+1=3,所以d=1,∵Sm=0,故ma1+m(m-1)2d=0,故a1=-(m-1∴am+am+1=2a1+(2m-1)d=-(m-1)+2m-1=5,解得∴a1=-5-12=-2,nSn=n-2n+n設(shè)f(n)=12n3-52n2,則f'(n)=32n2-令f'(n)=0,得n=103或n=由n∈N*,得當(dāng)n=3時(shí),nSn取最小值12×27-52×9=-913.16設(shè){an}的公差為d,由a12=38a5>0,得a1=-765d,a12<a5,即所以an=n-815d,從而可知當(dāng)1≤n≤16時(shí),a當(dāng)n≥17時(shí),an<0.從而b1>b2>…>b14>0>b17>b18>…,b15=a15a16a17<0,b16=a16故S14>S13>…>S1,S14>S15,S15<S16,S16>S17>S18>….因?yàn)閍15=-65d>0,a18=95d<0,所以a15+a18=-65d+95d=35d<0,所以b15+b16=a16a17(a15+a18)>0,所以S16>S14,故Sn中S14.解(1)∵數(shù)列{an}為等差數(shù)列,∴a3+a4=a2+a5=22.又a3·a4=117,∴a3,a4是方程x2-22x+117=0的兩實(shí)根.又公差d>0,∴a3<a4,∴a3=9,a4=13,∴∴通項(xiàng)公式an=4n-3.(2)由(1)知a1=1,d=4,∴Sn=na1+n(n-1)2d=∴當(dāng)n=1時(shí),Sn最小,最小值為S1=a1=1.(3)由(2)知Sn=2n2-n,∴bn=Sn∴b1=11+c,b2=62+c,∵數(shù)列{bn}是等差數(shù)列,∴2b2=b