教育单元训练金卷?高三?数学卷答案(A) 第十一单元 等差数列与等比数列
一、选择题(本大题共12小题,每小题5分,共60分,在每小题给出的四个选项中,只有一项是符合题目要求的) 1.【答案】D
【解析】根据题意可得,a7?a1?a7??a1?a1?6d?3?a1?2d?4,因为S7?2?72?56,
所以a1?3d?8,两式相减,得d?4,故选D. 2.【答案】D 【解析】∵S4?a1?a4?4?20,∴
2?20,∴a1?a4?10又a5?10,可得a1?a4?a5,a1?d,∴
a1?d?2,则a16?2??16?1??2?32,故选D.
3.【答案】A
【解析】由题得?a21?a13??a7?2a7?a7?3a7?2π,∴a7?3π,所以tana27?tan3???3, 故答案为A. 4.【答案】D
【解析】因为a3?a4?a11?18,所以可得3a1?15d?18?a1?5d?6, 所以S11?11?a1?5d??11?6?66,故选D. 5.【答案】C
【解析】∵a2?a21,a3,a4成等比数列,∴a31a4,即?a1?4??a1?a1?6?,解得a1??8, ∴a2?a3?2a1?6??10,故选C. 6.【答案】C
【解析】由题意,数列?an?为等比数列,且a2012?4,a2024?16,则a2018是a2012,a2024的等比中项,且是同号的,所以a2018?a2012?a2024?4?16?8,故选C. 7.【答案】A
【解析】因为a1a6?a2a5?a43a4?2,故a2a4?a,故选A. 3a58.【答案】C
a??1???2?41??【解析】由题意可得:
S41???2?S=?1?(﹣)22?5,故选C. 2a1??1???2?2??1???2?9.【答案】C
【解析】由题得a2?a6?14,a2a6?33,所以a2?3,a6?11或a2?11,a6?3, 当a2?3,a6?11时,d?11?36?2?2,a1?1,a7?13,∴a1a7?13, 当a11?32?11,a6?3时,d?6?2?2,a1?13,a7?1,∴a1a7?13,故答案为C. 10.【答案】B
【解析】设数列的公比为q,由题意可知:q?1,且2?a2?2??a1?1?a3, 即2??6?2??6q?1?6q,整理可得:2q2?5q?2?0,则q?2,(q?12舍去). 则a63??1?26?1?2?3,该数列的前6项和S6?1?2?189,故选B. 11.【答案】B
【解析】根据S13?0,S14?0,可以确定a1?a13?2a7?0,a1?a14?a7?a8?0, 所以可以得到a7?0,a8?0,所以则Sn取最大值时n的值为7,故选B. 12.【答案】D
【解析】在等比数列?an?中,由a6?a2?a10可得a1?25??a1?2???a1?29?, 解得a1?117?b1?b17?17??2b9?25,∴b9?2a7?2???1?25?26????4,∴S17?2?2?17b9?68,故选D.
二、填空题(本大题有4小题,每小题5分,共20分.请把答案填在题中横线上)
13.【答案】?1018081
【解析】由题意可得a2?a1??1,a4?a3??3,a6?a5??5,a2018?a2017??2017, 则数列?an?的前2019项的和为?1?3?L?2017??1?20172?1009??1018081. 14.【答案】a??4,n?1n? ?4n?1,n?2【解析】根据递推公式,可得S2n?1?2?n?1???n?1??1 由通项公式与求和公式的关系,可得an?Sn?Sn?1,代入化简得
a2n?2n2?n?1?2?n?1???n?1??1?4n?1,经检验,当n?1时,S1?4,a1?3
所以S?4,n?11?a1,所以an?? . ?4n?1,n?215.【答案】
613 【解析】∵等差数列?a13?a1?a13?13?2a76n?中S13?6,∴S13?2?2?6,∴a7?13,
设等差数列?a的公差为d,则3a6n?9?2a10?2?a9?a10??a9?a9?2d?a7?13.
16.【答案】40362019
【解析】由题意an?n?1?n?2,则
1a?2?1??2??1?n?1?n?1??, nn?n所以
1a?1?L?1?2?????1?1?????1?1???L???1?1?????1?4036??2?1????. 1a2a2018?2??23??20182019??20192019三、解答题(本大题有6小题,共70分.解答应写出文字说明、证明过程或演算步骤) 17.【答案】(1)3;(2)2或3. 【解析】(1)方法一:设?an?的公差为d,
由题,??a4?a1?3d?1?S15?15a1?105d?75 ,解得??a1??2?d?1 ,∴a6?a1?5d?3. 方法二:由题,S15?15a8?75,∴a8?5,于是aa4?a86?2?3. (2)方法一:Sn?n?1?n?na1?2d?n2?5n2,当n?2或3时,Sn取得最小值.
方法二:an?a1??n?1?d?n?3,∴a1?a2?a3?0?a4?L, 故当n?2或3时,Sn取得最小值.
18.【答案】(1)b?1n?2n;(2)q??4时,S3?18;q?3时,S3??3. 【解析】设等差数列?an?公差为d,等比数列?bn?公比为q?q?0?, 有??1?d??q?2,即d?q?3.
(1)∵??1?2d??q2?5,结合d?q?3得q?2,∴bn?2n?1. (2)∵T3?1?q?q2?13,解得q??4或3,
当q??4时,d?7,此时S3?a1?a2?a3??1?6?13?18;
当q?3时,d?0,此时S3?3a1??3. 19.【答案】(1)an?n?1?n?2n;(2)Tn?2.
【解析】(1)q?a2?a3a?a?2,a1?2,an?2n.
12(2)bn?n?1?n?log2an?n,Tn?2. 20.【答案】(1)an?n,n?N*;(2)T2n?A?B?22n?1?n?2,n?N*. 【解析】(1)当n?1时,a1?S1?1,
n2?n?n?1?2当n?2时,a?n?Sn?Sn?1?2??n?1?2?n. 当n?1时,a1?1也满足上式,由数列?an?的通项公式为a*n?n,n?N.
(2)由(1)知,bnn?2n???1??n,记数列?bn?的前2n项和T2n, 则T?22n?21?2?L?22n????1?2?3?4?L?2n?.
记A?21?22?L?22n,B??1?2?3?4?L?2n,则A?2?1?22n??11?2?22n?2,B???1?2????3?4??L?????2n?1??2n???n,
故数列?bn?的前2n项和T2n?A?B?22n?1?n?2,n?N*. 21.【答案】(1)an?2n;(2)Sn?nn?1. 【解析】(1)设数列?an?的公差为d,由a1?2和a2、a3、a4?1成等比数列,得?2+2d?2??2+d??3?3d?,解得d=2,或d=?1,
当d=?1时,a3?0,与a2、a3、a4?1成等比数列矛盾,舍去.∴d=2,
an?a1??n?1?d?2?2?n?1??2n即数列?an?的通项公式an?2n. (2)b22111n?n??a2??n?2n?2??n?n?1??n?n?1
n?所以Sn?a11111n1?a2?…?an?1?12?2?3?…?n?n?1?1?n?1?n?1. 22.【答案】(1)an?n;(2)Sn??n?1??2n?1. 【解析】(1)设等差数列?an?的公差为d.
23由题意可知a42?a2?a5?3?,∴?1?3d???1?d??4?4d?,解得d?1或d??,
5∵数列?an?单调递增,∴d?1,∴an?1?n?1?n. (2)由(1)可得bn?n?2n?1.
∴Sn?1?20?2?21?3?22?L?n?2n?1,① ∴2Sn?1?21?2?22?L??n?1??2n?1?n?2n,② ①②得?Sn?1?2?2?L?2∴Sn??n?1??2n?1.
12n?11?2n?n?2??n?2n??1??1?n??2n,
1?2n
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