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1、課時(shí)作業(yè)(十六)一、選擇題1(2020重慶卷,文)在等差數(shù)列an中,a1a910,則a5的值為()A5B6C8 D10答案A解析依題意得a1a92a510,a55,選A.2在等差數(shù)列an中,a2a6,則sin(2a4)()A. B.C D答案D解析a2a6,2a4,sin(2a4)sin()cos,選D.3(2020合肥質(zhì)檢)設(shè)Sn是等差數(shù)列an的前n項(xiàng)和,若a49,S315,則數(shù)列an的通項(xiàng)an()A2n3 B2n1C2n1 D2n3答案C解析由,所以通項(xiàng)an2n1.4等差數(shù)列an的前n項(xiàng)和為Sn,已知am1am12a0,S2m139,則m()A38 B39C20 D19答案C解析am1am
2、12a又am1am12amam1或0(舍去)S2m1(2m1)am(2m1)am39,2m139m20.5設(shè)an是公差為正數(shù)的等差數(shù)列,若a1a2a315,a1a2a380,則a11a12a13()A120 B105C90 D75答案B解析設(shè)公差為d且d0.由已知,得.解得a12,d3(d0)a11a12a133a123(a111d)1056若兩個(gè)等差數(shù)列an和bn的前n項(xiàng)和分別是Sn,Tn,已知,則等于()A7 B.C. D.答案D解析.7等差數(shù)列an的前n項(xiàng)和為Sn,且S36,a34,則公差d等于()A1 B.C2 D3答案C解析由,解得d2.二、填空題8已知Sn是等差數(shù)列an的前n項(xiàng)和,
3、且a415,S555,則過(guò)點(diǎn)P(3,a3)、Q(4,a4)的直線的斜率是_解析設(shè)數(shù)列an的公差為d,則依題意,得,故直線PQ的斜率為4.9已知數(shù)列an中,a32,a51,若是等差數(shù)列,則a11_.答案0解析記bn,則b3,b5,數(shù)列bn的公差為(),b1,bn,即,an,故a110.10等差數(shù)列an中,Sn是其前n項(xiàng)和,a12020,2,則S2020的值為_(kāi)答案2020解析在等差數(shù)列an中,設(shè)公差為d,則a1(n1),a12020a12020d2,S202020202020220202020202020202020.11方程(x2xm)(x2xn)0有四個(gè)不等實(shí)根,且組成一個(gè)公差為的等差數(shù)列,
4、則mn的值為_(kāi)答案解析設(shè)四個(gè)根組成的等差數(shù)列為x1,x2,x3,x4,根據(jù)等差數(shù)列的性質(zhì),則有x1x4x2x312x13d1,又d,x1x2,x3,x4mn(x1x4)(x2x3)12(2020浙江卷,文)在如下數(shù)表中,已知每行、每列中的數(shù)都成等差數(shù)列,那么位于表中的第n行第n1列的數(shù)是_答案n2n解析第n行的第一個(gè)數(shù)是n,第n行的數(shù)構(gòu)成以n為公差的等差數(shù)列,則其第n1項(xiàng)為nnnn2n.13(2020蘇北四市調(diào)研)已知數(shù)列an共有m項(xiàng),記an的所有項(xiàng)和為S(1),第二項(xiàng)及以后所有項(xiàng)和為S(2),第三項(xiàng)及以后所有項(xiàng)和為S(3),第n項(xiàng)及以后所有項(xiàng)和為S(n),若S(n)是首項(xiàng)為1,公差為2的等差數(shù)列的前n項(xiàng)和,則當(dāng)nm時(shí),an_.答案2n1解析由題意得S(n)anamn12n2,當(dāng)n0,S1477,求所有可能的數(shù)列an的通項(xiàng)公式答案(1)an222n(2)an12n和an13n解(1)由S1498得2a113d14,又a11a110d0,故解得d2,a120.因此an的通項(xiàng)公式是an222n,n1,2,3,.(2)由,得,即由得7d.由得13d1,即d.于是d.又dZ,故d1.將代入得10a112.又a1Z,故a111或a112.所以所有可能的數(shù)列an的通項(xiàng)公式是an12n和an13n,n1,2,3,.