（福建專版）2019高考數(shù)學一輪復習 課時規(guī)范練28 等差數(shù)列及其前n項和 文.docx

課時規(guī)范練28等差數(shù)列及其前n項和基礎鞏固組1.已知等差數(shù)列an中,a4+a5=a3,a7=-2,則a9=() A.-8B.-6C.-4D.-22.(2017陜西咸陽二模)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)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.3B.4C.5D.63.已知在每項均大于零的數(shù)列an中,首項a1=1,且前n項和Sn滿足SnSn-1-Sn-1Sn=2SnSn-1(nN*,且n2),則a81等于()A.638B.639C.640D.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項和為Sn,則使得Sn取最大值時,n的值是()A.18B.19C.20D.215.(2017遼寧沈陽質(zhì)量檢測)設Sn為等差數(shù)列an的前n項和,若a1=1,公差d=2,Sn+2-Sn=36,則n=()A.5B.6C.7D.86.(2017北京豐臺一模)已知an為等差數(shù)列,Sn為其前n項和.若a2=2,S9=9,則a8=.7.已知在數(shù)列an中,a1=1,a2=2,當整數(shù)n2時,Sn+1+Sn-1=2(Sn+S1)都成立,則S15=.8.一個等差數(shù)列的前12項的和為354,前12項中偶數(shù)項的和與奇數(shù)項的和的比為3227,則該數(shù)列的公差d=.9.若數(shù)列an的前n項和為Sn,且滿足an+2SnSn-1=0(n2),a1=12.(1)求證:1Sn成等差數(shù)列;(2)求數(shù)列an的通項公式.導學號2419091110.(2017北京海淀一模,文15)已知等差數(shù)列an滿足a1+a2=6,a2+a3=10.(1)求數(shù)列an的通項公式;(2)求數(shù)列an+an+1的前n項和.導學號24190912綜合提升組11.若數(shù)列an滿足:a1=19,an+1=an-3(nN*),則數(shù)列an的前n項和數(shù)值最大時,n的值為()A.6B.7C.8D.912.(2017四川廣元二診)設等差數(shù)列an的前n項和為Sn,若Sm-1=-2,Sm=0,Sm+1=3,其中m2,則nSn的最小值為()A.-3B.-5C.-6D.-913.數(shù)列an是等差數(shù)列,數(shù)列bn滿足bn=anan+1an+2(nN*),設Sn為bn的前n項和.若a12=38a5>0,則當Sn取得最大值時,n的值等于.14.已知公差大于零的等差數(shù)列an的前n項和為Sn,且滿足a3a4=117,a2+a5=22.(1)求通項公式an;(2)求Sn的最小值;(3)若數(shù)列bn是等差數(shù)列,且bn=Snn+c,求非零常數(shù)c.導學號24190913創(chuàng)新應用組15.有兩個等差數(shù)列an,bn,其前n項和分別為Sn和Tn,若SnTn=3n2n+1,則a1+a2+a14+a19b1+b3+b17+b19的值為()A.2719B.1813C.107D.1713導學號24190914答案：1.B解法一:由已知可得2a1+7d=a1+2d,a1+6d=-2,解得a1=10,d=-2,所以a9=10+(-2)8=-6.解法二:因為a4+a5=a3,所以a3+a6=a3,a6=0,又a7=-2,所以d=-2,a9=-2+(-2)2=-6.2.C設第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C.3.C由已知SnSn-1-Sn-1Sn=2SnSn-1,可得Sn-Sn-1=2,Sn是以1為首項,2為公差的等差數(shù)列,故Sn=2n-1,Sn=(2n-1)2,a81=S81-S80=1612-1592=640,故選C.4.Ca1+a3+a5=105a3=35,a2+a4+a6=99a4=33,則an的公差d=33-35=-2,a1=a3-2d=39,Sn=-n2+40n,因此當Sn取得最大值時,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+4=36,所以n=8.解法二:Sn+2-Sn=an+1+an+2=2a1+(2n+1)d=2+2(2n+1)=36,解得n=8.6.0an為等差數(shù)列,Sn為其前n項和,a2=2,S9=9,a2=a1+d=2,S9=9a1+982d=9,解得d=-13,a1=73,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(n2),數(shù)列an從第二項起構成以2為首項,2為公差的等差數(shù)列,則S15=1+214+141322=211.8.5設該等差數(shù)列的前12項中奇數(shù)項的和為S奇,偶數(shù)項的和為S偶,等差數(shù)列的公差為d.由題意得S奇+S偶=354,S偶S奇=3227,解得S偶=192,S奇=162.又S偶-S奇=6d,所以d=192-1626=5.9.(1)證明 當n2時,由an+2SnSn-1=0,得Sn-Sn-1=-2SnSn-1,所以1Sn-1Sn-1=2.又1S1=1a1=2,故1Sn是首項為2,公差為2的等差數(shù)列.(2)解 由(1)可得1Sn=2n,Sn=12n.當n2時,an=Sn-Sn-1=12n-12(n-1)=n-1-n2n(n-1)=-12n(n-1).當n=1時,a1=12不適合上式.故an=12,n=1,-12n(n-1),n2.10.解 (1)設數(shù)列an的公差為d,a1+a2=6,a2+a3=10,a3-a1=4,2d=4,d=2.又a1+a1+d=6,a1=2,an=a1+(n-1)d=2n.(2)記bn=an+an+1,則bn=2n+2(n+1)=4n+2,又bn+1-bn=4(n+1)+2-4n-2=4,bn是首項為6,公差為4的等差數(shù)列,其前n項和Sn=n(b1+bn)2=n(6+4n+2)2=2n2+4n.11.Ba1=19,an+1-an=-3,數(shù)列an是以19為首項,-3為公差的等差數(shù)列.an=19+(n-1)(-3)=22-3n.設an的前k項和數(shù)值最大,則有ak0,ak+10,kN*.22-3k0,22-3(k+1)0.193k223.kN*,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)2,am+am+1=5,am+am+1=2a1+(2m-1)d=-(m-1)+2m-1=5,解得m=5.a1=-5-12=-2,nSn=n-2n+n(n-1)21=12n3-52n2,設f(n)=12n3-52n2,則f(n)=32n2-5n,令f(n)=0,得n=103或n=0,由nN*,得當n=3時,nSn取最小值1227-529=-9.故選D.13.16設an的公差為d,由a12=38a5>0,得a1=-765d,a12<a5,即d<0,所以an=n-815d,從而可知當1n16時,an>0;當n17時,an<0.從而b1>b2>>b14>0>b17>b18>,b15=a15a16a17<0,b16=a16a17a18>0,故S14>S13>>S1,S14>S15,S15<S16,S16>S17>S18>.因為a15=-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S16最大.故答案為16.14.解 (1)數(shù)列an為等差數(shù)列,a3+a4=a2+a5=22.又a3a4=117,a3,a4是方程x2-22x+117=0的兩實根.又公差d>0,a3<a4,a3=9,a4=13,a1+2d=9,a1+3d=13,a1=1,d=4.通項公式an=4n-3.(2)由(1)知a1=1,d=4,Sn=na1+n(n-1)2d=2n2-n=2n-142-18.當n=1時,Sn最小,最小值為S1=a1=1.(3)由(2)知Sn=2n2-n,bn=Snn+c=2n2-nn+c,b1=11+c,b2=62+c,b3=153+c.數(shù)列bn是等差數(shù)列,2b2=b1+b3,即62+c2=11+c+153+c,2c2+c=0,c=-12(c=0舍去),故c=-12.15.D由題意,知a1+a2+a14+a19=2(a8+a10)=4a9,同理b1+b3+b17+b19=4b10,又SnTn=3n2n+1,且Sn和Tn都是關于n的二次函數(shù),設Sn=kn3n=3kn2,設Tn=kn(2n+1),a9=S9-S8=3k17,b10=T10-T9=39k,a1+a2+a14+a19b1+b3+b17+b19=a9b10=3k1739k=1713.