鲁京津琼专用2020版高考数学大一轮复习第六章数列阶段自测卷四课件
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1、阶段自测卷(四),第六章 数列,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22,1.(2019衡水中学考试)已知等差数列an的公差为2,前n项和为Sn,且S10100,则a7的值为 A.11 B.12 C.13 D.14,所以a11. 所以an2n1, 故a713.故选C.,一、选择题(本大题共12小题,每小题5分,共60分),1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22,解析 设等差数列的首项为a1,公差为d, 则a3a12d,a7a16d. 因为a1,a3,
2、a7成等比数列, 所以(a12d)2a1(a16d),,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22,3.(2019四省联考)已知等差数列an的前n项和为Sn,若S630,S1010,则S16等于 A.160 B.80 C.20 D.40,解得a110,d2, 故S1616a1120d1610120(2)80,故选B.,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22,A.3 B.5 C.31 D.33,1,2,3,4,5,6,7,8,9,10,11,12,13,
3、14,15,16,17,18,19,20,21,22,5.(2019湖南五市十校联考)已知数列an满足2anan1an1(n2),a2a4 a612,a1a3a59,则a1a6等于 A.6 B.7 C.8 D.9,解析 由数列an满足2anan1an1(n2)得数列an为等差数列, 所以a2a4a63a412,即a44, 同理a1a3a53a39,即a33, 所以a1a6a3a47.,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22,解析 依题意可知,这个同学第1天,第2天,跑的路程依次成首项为5 000,公差为200的等差数列,
4、,6.(2019新乡模拟)为了参加冬季运动会的5 000 m长跑比赛,某同学给自己制定了7天的训练计划:第1天跑5 000 m,以后每天比前1天多跑200 m,则这个同学7天一共将跑 A.39 200 m B.39 300 m C.39 400 m D.39 500 m,A.38 B.20 C.10 D.9,解析 因为an是等差数列,所以am1am12am,,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22,由S2m138知am0,所以am2,,即(2m1)238,解得m10,故选C.,1,2,3,4,5,6,7,8,9,10,1
5、1,12,13,14,15,16,17,18,19,20,21,22,解析 设等比数列an的公比为q, 3a2 , 2a3,a4成等差数列, 22a33a2a4, 4a2q3a2a2q2,化为q24q30, 解得q1或3. 又数列的各项均不相等,q1,,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22,9.(2019广东六校联考)将正奇数数列1,3,5,7,9,依次按两项、三项分组,得到分组序列如下: (1,3),(5,7,9),(11,13),(15,17,19),称(1,3)为第1组,(5,7,9)为第2组,依此类推,则原数列
6、中的2 019位于分组序列中的 A.第404组 B.第405组 C.第808组 D.第809组,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22,解析 正奇数数列1,3,5,7,9,的通项公式为an2n1, 则2 019为第1 010个奇数, 因为按两项、三项分组, 故按5个一组分组是有202组, 故原数列中的2 019位于分组序列中的第404组,故选A.,1,2,3,4,5,6,7,8,9,10,11,12,13,1
7、4,15,16,17,18,19,20,21,22,A.1 290 B.1 280 C.1 281 D.1 821,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22,当n2时,anSnSn1(n1)2n21, 故 a91012811 281.,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22,解析 由Snn24n,可得an2n3,,1,2,3,4,5,6,
8、7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22,二、填空题(本大题共4小题,每小题5分,共20分),1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22,13.设等差数列an的公差为d,其前n项和为Sn,若a4a100,2S12S210,则d的值为_.,10,解析 由a
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