1、基础达标1(2022河北省唐山质量检测)已知等差数列an的前n项和为Sn,a415,S555,则数列an的公差是()A B4C4 D3解析:选Ban是等差数列,a415,S555,a1a522,2a322,a311,公差da4a34.2(2022辽宁大连市双基测试)设an是公差为正数的等差数列,若a1a2a515,且a1a2a380,则a11a12a13等于()A120 B105C90 D75解析:选B设数列an的公差为d,由a1a2a315,得a25,由a1a2a380,得a1a3(5d)(5d)16,故25d216,d3,则a12,a11a12a133a133d699105.3(2022安
2、徽望江中学第九次模拟)设数列an是公差d0的等差数列an的四个命题:p1:数列an是递增数列;p2:数列nan是递增数列;p3:数列是递增数列;p4:数列an3nd是递增数列其中的真命题为()Ap1,p2 Bp3,p4Cp2,p3 Dp1,p4解析:选D由于d0,所以an1an,所以p1是真命题由于n1n,但是an的符号不知道,所以p2是假命题同理p3是假命题由an13(n1)dan3nd4d0,所以p4是真命题5(2022浙江省名校联考)已知每项均大于零的数列an中,首项a11且前n项和Sn满足SnSn1 2(nN*且n2),则a81()A638 B639C640 D641解析:选C由已知S
3、nSn1 2可得,2,是以1为首项,2为公差的等差数列,故2n1,Sn(2n1)2,a81S81S8016121592640.6(2021高考广东卷)在等差数列an中,已知a3a810,则3a5a7_解析:法一:a3a82a19d10,3a5a74a118d2(2a19d)21020.法二:a3a82a35d10,3a5a74a310d2(2a35d)21020.答案:207在数列an中,若点(n,an)在经过点(5,3)的定直线l上,则数列an的前9项和S9_解析:点(n,an)在定直线l上,数列an为等差数列,ana1(n1)D将(5,3)代入,得3a14da5.S9(a1a9)9a539
4、27.答案:278(2022河南三市调研)设数列an的通项公式为an2n10(nN*),则|a1|a2|a15|_解析:由an2n10(nN*)知an是以8为首项,2为公差的等差数列,又由an2n100得n5,当n5时,an0,当n5时,an0,|a1|a2|a15|(a1a2a3a4)(a5a6a15)20110130.答案:1309(2022浙江温州市适应性测试)已知an是递增的等差数列,a12,aa48.(1)求数列an的通项公式;(2)若bnan2an,求数列bn的前n项和Sn.解:(1)设等差数列的公差为d,d0.由题意得,(2d)223d8,d2d6(d3)(d2)0,得d2.故a
5、na1(n1)d2(n1)22n,得an2n.(2)bnan2an2n22n.Snb1b2bn(222)(424)(2n22n)(2462n)(222422n)n(n1).10已知数列an的各项均为正数,前n项和为Sn,且满足2Snan4(nN*)(1)求证:数列an为等差数列;(2)求数列an的通项公式解:(1)证明:当n1时,有2a1a14,即a2a130,解得a13(a11舍去)当n2时,有2Sn1an5,又2Snan4,两式相减得2anaa1,即a2an1a,也即(an1)2a,因此an1an1或an1an1.若an1an1,则anan11.而a13,所以a22,这与数列an的各项均为
6、正数相冲突,所以an1an1,即anan11,因此数列an为等差数列(2)由(1)知a13,d1,所以数列an的通项公式an3(n1)1n2,即ann2.力气提升1数列an满足a11,an1ranr(nN*,rR且r0),则“r1”是“数列an为等差数列”的()A充分不必要条件B必要不充分条件C充分必要条件D既不充分也不必要条件解析:选A当r1时,易知数列an为等差数列;由题意易知a22r,a32r2r,当数列an是等差数列时,a2a1a3a2,即2r12r2r,解得r或r1,故“r1”是“数列an为等差数列”的充分不必要条件2已知Sn是等差数列an的前n项和,S100并且S110,若SnSk
7、对nN*恒成立,则正整数k构成的集合为()A5 B6C5,6 D7解析:选C在等差数列an中,由S100,S110得,S100a1a100a5a60,S110a1a112a60,故可知等差数列an是递减数列且a60,所以S5S6Sn,其中nN*,所以k5或6.3(2022湖北荆门调研)已知一等差数列的前四项和为124,后四项和为156,各项和为210,则此等差数列的项数是_解析:设数列an为该等差数列,依题意得a1an70.Sn210,Sn,210,n6.答案:64(2022福建龙岩质检)已知数列an的首项为2,数列bn为等差数列且bnan1an(nN*)若b22,b78,则a8_解析:bn为
8、等差数列,且b22,b78,设其公差为d,b7b25d,即825Dd2.bn2(n2)22n6.an1an2n6.由a2a1216,a3a2226,anan12(n1)6,累加得:ana12(12n1)6(n1)n27n6,ann27n8.a816.答案:165(2022山东济南模拟)设同时满足条件:bn1(nN*);bnM(nN*,M是与n无关的常数)的无穷数列bn叫“特界”数列(1)若数列an为等差数列,Sn是其前n项和,a34,S318,求Sn;(2)推断(1)中的数列Sn是否为“特界”数列,并说明理由解:(1)设等差数列an的公差为d,则a12d4,S3a1a2a33a13d18,解得
9、a18,d2,Snna1dn29n.(2)由Sn110,得Sn1,故数列Sn适合条件.而Snn29n(nN*),则当n4或5时,Sn有最大值20,即Sn20,故数列Sn适合条件.综上,数列Sn是“特界”数列6(选做题)(2022广东深圳质检)各项均为正数的数列an满足a4Sn2an1(nN*),其中Sn为an的前n项和(1)求a1,a2的值;(2)求数列an的通项公式;(3)是否存在正整数m、n,使得向量a(2an2,m)与向量b(an5,3an)垂直?说明理由解:(1)当n1时,a4S12a11,即(a11)20,解得a11.当n2时,a4S22a214a12a2132a2,解得a23或a21(舍去)(2)a4Sn2an1,a4Sn12an11.得:aa4an12an12an2(an1an),即(an1an)(an1an)2(an1an)数列an各项均为正数,an1an0,an1an2,数列an是首项为1,公差为2的等差数列an2n1.(3)an2n1,a(2an2,m)(2(2n3),m)0,b(an5,3an)(2n9),2(n1)0,abab0m(n1)(2n3)(2n9)2(n1)12(n1)7m(n1)4(n1)216(n1)7m4(n1)16.m,nN*,n17,m47161,即n6,m45.当n6,m45时,aB