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新版高中数学人教A版必修5习题：第二章数列 2-3-1 含解析-2019最新整理

第1课时 等差数列的前n项和

课时过关·能力提升 基础巩固

1等差数列{an}的前n项和为Sn,若a1=2,S3=12,则a6等于( ). A.8 答案:C

2数列{an}的前n项和为Sn,若Sn=2n2-18n,则当Sn取得最小值时,n的值为( ). A.4或5 答案:A

3设数列{an}的前n项和Sn=n2,则a8的值为( ). A.15

B.16

C.49

D.64

talk with her4. sleep well5. good for sleepB.10 C.12 D.14

B.5或6 C.4 D.5

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Maybe you could get a part-time job in the evening.

解析:a8=S8-S7=64-49=15. 答案:A

4已知等差数列{an}的前n项和为Sn,若a4=18-a5,则S8等于( ). A.18

B.36

C.54

D.72

解析:∵a4=18-a5,∴a4+a5=18.

∴S8答案:D

5在等差数列{an}中,首项a1=0,公差d≠0,若ak=a1+a2+a3+…+a7,则k等于( ). A.21

B.22

C.23

D.24

解析:由题意得ak=a1+(k-1)d=(k-1)d,

a1+a2+a3+…+a7=21d,

所以(k-1)d=21d.

又d≠0,所以k-1=21,所以k=22. 答案:B

talk with her4. sleep well5. good for sleep

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Maybe you could get a part-time job in the evening.

6已知等差数列{an}的前n项和为Sn,若a2=1,S5=10,则S7= .

解析:由S5

得a3=2,故a4=3,S7答案:21

7已知数列{an}的前n项和Sn=2n-3,则数列{an}的通项公式为 .

解析:当n=1时,a1=S1=21-3=-1;

当n≥2时,an=Sn-Sn-1=2n-3-2n-1+3=2n-1.

又a1=-1不满足上式,故an答案:an

8等差数列{an}的前n项和为Sn,且S2 016=a2 016=2 016,则a1= .

解析:S2016016,解得a1=-2014.答案:-2 014

talk with her4. sleep well5. good for sleep

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