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课时规范练28等差数列及其前n项和基础巩固组1.已知等差数列an中,a4+a5=a3,a7=-2,则a9=() A.-8B.-6C.-4D.-22.(2017陕西咸阳二模)张丘建算经卷上一题为“今有女善织,日益功疾,且从第二天起,每天比前一天多织相同量的布,现在一月(按30天计)共织布390尺,最后一天织布21尺”,则该女第一天织布多少尺?()A.3B.4C.5D.63.已知在每项均大于零的数列an中,首项a1=1,且前n项和Sn满足SnSn-1-Sn-1Sn=2SnSn-1(nN*,且n2),则a81等于()A.638B.639C.640D.6414.已知数列an是等差数列,a1+a3+a5=105,a2+a4+a6=99,an的前n项和为Sn,则使得Sn取最大值时,n的值是()A.18B.19C.20D.215.(2017辽宁沈阳质量检测)设Sn为等差数列an的前n项和,若a1=1,公差d=2,Sn+2-Sn=36,则n=()A.5B.6C.7D.86.(2017北京丰台一模)已知an为等差数列,Sn为其前n项和.若a2=2,S9=9,则a8=.7.已知在数列an中,a1=1,a2=2,当整数n2时,Sn+1+Sn-1=2(Sn+S1)都成立,则S15=.8.一个等差数列的前12项的和为354,前12项中偶数项的和与奇数项的和的比为3227,则该数列的公差d=.9.若数列an的前n项和为Sn,且满足an+2SnSn-1=0(n2),a1=12.(1)求证:1Sn成等差数列;(2)求数列an的通项公式.导学号2419091110.(2017北京海淀一模,文15)已知等差数列an满足a1+a2=6,a2+a3=10.(1)求数列an的通项公式;(2)求数列an+an+1的前n项和.导学号24190912综合提升组11.若数列an满足:a1=19,an+1=an-3(nN*),则数列an的前n项和数值最大时,n的值为()A.6B.7C.8D.912.(2017四川广元二诊)设等差数列an的前n项和为Sn,若Sm-1=-2,Sm=0,Sm+1=3,其中m2,则nSn的最小值为()A.-3B.-5C.-6D.-913.数列an是等差数列,数列bn满足bn=anan+1an+2(nN*),设Sn为bn的前n项和.若a12=38a50,则当Sn取得最大值时,n的值等于.14.已知公差大于零的等差数列an的前n项和为Sn,且满足a3a4=117,a2+a5=22.(1)求通项公式an;(2)求Sn的最小值;(3)若数列bn是等差数列,且bn=Snn+c,求非零常数c.导学号24190913创新应用组15.有两个等差数列an,bn,其前n项和分别为Sn和Tn,若SnTn=3n2n+1,则a1+a2+a14+a19b1+b3+b17+b19的值为()A.2719B.1813C.107D.1713导学号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:因为a4+a5=a3,所以a3+a6=a3,a6=0,又a7=-2,所以d=-2,a9=-2+(-2)2=-6.2.C设第n天织布an尺,则数列an是等差数列,且S30=390,a30=21,S30=302(a1+a30),即390=15(a1+21),解得a1=5.故选C.3.C由已知SnSn-1-Sn-1Sn=2SnSn-1,可得Sn-Sn-1=2,Sn是以1为首项,2为公差的等差数列,故Sn=2n-1,Sn=(2n-1)2,a81=S81-S80=1612-1592=640,故选C.4.Ca1+a3+a5=105a3=35,a2+a4+a6=99a4=33,则an的公差d=33-35=-2,a1=a3-2d=39,Sn=-n2+40n,因此当Sn取得最大值时,n=20.5.D解法一:由题知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为等差数列,Sn为其前n项和,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),数列an从第二项起构成以2为首项,2为公差的等差数列,则S15=1+214+141322=211.8.5设该等差数列的前12项中奇数项的和为S奇,偶数项的和为S偶,等差数列的公差为d.由题意得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)证明 当n2时,由an+2SnSn-1=0,得Sn-Sn-1=-2SnSn-1,所以1Sn-1Sn-1=2.又1S1=1a1=2,故1Sn是首项为2,公差为2的等差数列.(2)解 由(1)可得1Sn=2n,Sn=12n.当n2时,an=Sn-Sn-1=12n-12(n-1)=n-1-n2n(n-1)=-12n(n-1).当n=1时,a1=12不适合上式.故an=12,n=1,-12n(n-1),n2.10.解 (1)设数列an的公差为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)记bn=an+an+1,则bn=2n+2(n+1)=4n+2,又bn+1-bn=4(n+1)+2-4n-2=4,bn是首项为6,公差为4的等差数列,其前n项和Sn=n(b1+bn)2=n(6+4n+2)2=2n2+4n.11.Ba1=19,an+1-an=-3,数列an是以19为首项,-3为公差的等差数列.an=19+(n-1)(-3)=22-3n.设an的前k项和数值最大,则有ak0,ak+10,kN*.22-3k0,22-3(k+1)0.193k223.kN*,k=7.满足条件的n的值为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,设f(n)=12n3-52n2,则f(n)=32n2-5n,令f(n)=0,得n=103或n=0,由nN*,得当n=3时,nSn取最小值1227-529=-9.故选D.13.16设an的公差为d,由a12=38a50,得a1=-765d,a12a5,即d0;当n17时,anb2b140b17b18,b15=a15a16a170,故S14S13S1,S14S15,S15S17S18.因为a15=-65d0,a18=95d0,所以a15+a18=-65d+95d=35d0,所以S16S14,故Sn中S16最大.故答案为16.14.解 (1)数列an为等差数列,a3+a4=a2+a5=22.又a3a4=117,a3,a4是方程x2-22x+117=0的两实根.又公差d0,a3a4,a3=9,a4=13,a1+2d=9,a1+3d=13,a1=1,d=4.通项公式an=4n-3.(2)由(1)知a1=1,d=4,Sn=na1+n(n-1)2d=2n2-n=2n-142-18.当n=1时,Sn最小,最小值为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.数列bn是等差数列,2b2=b1+b3,即62+c2=11+c+153+c,2c2+c=0,c=-12(c=0舍去),故c=-12.15.D由题意,知a1+a2+a14+a19=2(a8+a10)=4a9,同理b1+b3+b17+b19=4b10,又SnTn=3n2n+1,且Sn和Tn都是关于n的二次函数,设Sn=kn3n=3kn2,设Tn=kn(2n+1),a9=S9-S8=3k17,b10=T10-T9=39k,a1+a2+a14+a19b1+b3+b17+b19=a9b10=3k1739k=1713.
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