6.2等差数列典型例题及详细解答.pdf

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1、1等差数列的定义一般地，如果一个数列从第2 项起，每一项与它的前一项的差等于同一个常数，那么这个数列就叫做等差数列，这个常数叫做等差数列的公差，通常用字母_d_表示2等差数列的通项公式如果等差数列an 的首项为a1，公差为d，那么它的通项公式是ana1(n1)d.3等差中项如果Aab2，那么A叫做a与b的等差中项4等差数列的常用性质(1)通项公式的推广：anam(nm)d(n，mN*)(2)若an 为等差数列，且klmn(k，l，m，nN*)，则akalaman.(3)若an 是等差数列，公差为d，则 a2n 也是等差数列，公差为2d.(4)若an，bn是等差数列，则panqbn也是等差数列(

2、5)若an 是等差数列，公差为d，则ak，akm，ak2m，(k，mN*)是公差为md的等差数列5等差数列的前n项和公式设等差数列 an的公差为d，其前n项和Snna1an2或Snna1nn12d.6等差数列的前n项和公式与函数的关系Snd2n2a1d2n.数列 an是等差数列?SnAn2Bn(A、B为常数)7等差数列的前n项和的最值在等差数列 an中，a10，d0，则Sn存在最 _大_值；若a10，则Sn存在最 _小_值【思考辨析】判断下面结论是否正确(请在括号中打“”或“”)(1)若一个数列从第二项起每一项与它的前一项的差都是常数，则这个数列是等差数列()(2)数列 an 为等差数列的充要

3、条件是对任意nN*，都有 2an1anan2.()(3)等差数列 an的单调性是由公差d决定的()(4)数列 an 为等差数列的充要条件是其通项公式为n的一次函数()(5)数列 an 满足an1ann，则数列 an 是等差数列()(6)已知数列 an 的通项公式是anpnq(其中p，q为常数)，则数列 an 一定是等差数列()1(2015重庆)在等差数列 an中，若a24，a42，则a6等于()A 1 B 0 C 1 D 6答案B解析由等差数列的性质，得a62a4a222 40，选 B.2(2014福建)等差数列 an的前n项和为Sn，若a12，S3 12，则a6等于()A8 B 10 C 1

4、2 D 14答案C解析由题意知a12，由S3 3a1322d12，解得d2，所以a6a15d252 12，故选 C.3在等差数列an 中，已知a4a816，则该数列前11 项和S11等于()A58 B 88 C 143 D 176答案B解析S1111a1a11211a4a8288.4设数列 an是等差数列，若a3a4a5 12，则a1a2a7等于()A14 B 21 C 28 D 35答案C解析a3a4a53a412，a44，a1a2a7 7a428.5(2014北京)若等差数列 an满足a7a8a90，a7a100，则当n_时，an的前n项和最大答案8解析因为数列 an是等差数列，且a7a8

5、a93a80，所以a80.又a7a10a8a90，所以a9 0.故当n8 时，其前n项和最大题型一等差数列基本量的运算文档编码:CZ7G2C7D10P8 HE5I10G8M6L9 ZU10E6P8I3I2文档编码:CZ7G2C7D10P8 HE5I10G8M6L9 ZU10E6P8I3I2文档编码:CZ7G2C7D10P8 HE5I10G8M6L9 ZU10E6P8I3I2文档编码:CZ7G2C7D10P8 HE5I10G8M6L9 ZU10E6P8I3I2文档编码:CZ7G2C7D10P8 HE5I10G8M6L9 ZU10E6P8I3I2文档编码:CZ7G2C7D10P8 HE5I10G8

6、M6L9 ZU10E6P8I3I2文档编码:CZ7G2C7D10P8 HE5I10G8M6L9 ZU10E6P8I3I2文档编码:CZ7G2C7D10P8 HE5I10G8M6L9 ZU10E6P8I3I2文档编码:CZ7G2C7D10P8 HE5I10G8M6L9 ZU10E6P8I3I2文档编码:CZ7G2C7D10P8 HE5I10G8M6L9 ZU10E6P8I3I2文档编码:CZ7G2C7D10P8 HE5I10G8M6L9 ZU10E6P8I3I2文档编码:CZ7G2C7D10P8 HE5I10G8M6L9 ZU10E6P8I3I2文档编码:CZ7G2C7D10P8 HE5I10G

7、8M6L9 ZU10E6P8I3I2文档编码:CZ7G2C7D10P8 HE5I10G8M6L9 ZU10E6P8I3I2文档编码:CZ7G2C7D10P8 HE5I10G8M6L9 ZU10E6P8I3I2文档编码:CZ7G2C7D10P8 HE5I10G8M6L9 ZU10E6P8I3I2文档编码:CZ7G2C7D10P8 HE5I10G8M6L9 ZU10E6P8I3I2文档编码:CZ7G2C7D10P8 HE5I10G8M6L9 ZU10E6P8I3I2文档编码:CZ7G2C7D10P8 HE5I10G8M6L9 ZU10E6P8I3I2文档编码:CZ7G2C7D10P8 HE5I10

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9、0G8M6L9 ZU10E6P8I3I2文档编码:CZ7G2C7D10P8 HE5I10G8M6L9 ZU10E6P8I3I2文档编码:CZ7G2C7D10P8 HE5I10G8M6L9 ZU10E6P8I3I2文档编码:CZ7G2C7D10P8 HE5I10G8M6L9 ZU10E6P8I3I2文档编码:CZ7G2C7D10P8 HE5I10G8M6L9 ZU10E6P8I3I2文档编码:CZ7G2C7D10P8 HE5I10G8M6L9 ZU10E6P8I3I2文档编码:CZ7G2C7D10P8 HE5I10G8M6L9 ZU10E6P8I3I2文档编码:CZ7G2C7D10P8 HE5I

10、10G8M6L9 ZU10E6P8I3I2文档编码:CZ7G2C7D10P8 HE5I10G8M6L9 ZU10E6P8I3I2文档编码:CZ7G2C7D10P8 HE5I10G8M6L9 ZU10E6P8I3I2文档编码:CZ7G2C7D10P8 HE5I10G8M6L9 ZU10E6P8I3I2文档编码:CZ7G2C7D10P8 HE5I10G8M6L9 ZU10E6P8I3I2文档编码:CZ7G2C7D10P8 HE5I10G8M6L9 ZU10E6P8I3I2文档编码:CZ7G2C7D10P8 HE5I10G8M6L9 ZU10E6P8I3I2文档编码:CZ7G2C7D10P8 HE5

11、I10G8M6L9 ZU10E6P8I3I2文档编码:CZ7G2C7D10P8 HE5I10G8M6L9 ZU10E6P8I3I2文档编码:CZ7G2C7D10P8 HE5I10G8M6L9 ZU10E6P8I3I2文档编码:CZ7G2C7D10P8 HE5I10G8M6L9 ZU10E6P8I3I2文档编码:CZ7G2C7D10P8 HE5I10G8M6L9 ZU10E6P8I3I2文档编码:CZ7G2C7D10P8 HE5I10G8M6L9 ZU10E6P8I3I2文档编码:CZ7G2C7D10P8 HE5I10G8M6L9 ZU10E6P8I3I2文档编码:CZ7G2C7D10P8 HE

12、5I10G8M6L9 ZU10E6P8I3I2例 1(1)在数列 an中，若a1 2，且对任意的nN*有 2an112an，则数列 an 前 10 项的和为()A2 B 10 (2)已知在等差数列an 中，a27，a415，则前 10 项和S10等于()A100 B210C380 D400答案(1)C(2)B解析(1)由 2an 112an得an1an12，所以数列 an 是首项为 2，公差为12的等差数列，所以S1010(2)1010121252.(2)因为a27，a415，所以d4，a13，故S10103121094 210.思维升华(1)等差数列运算问题的一般求法是设出首项a1和公差d，

13、然后由通项公式或前n项和公式转化为方程(组)求解(2)等差数列的通项公式及前n项和公式，共涉及五个量a1，an，d，n，Sn，知其中三个就能求另外两个，体现了方程的思想(1)(2015课标全国)设Sn是等差数列 an的前n项和，若a1a3a53，则S5等于()A5 B 7 C 9 D 11(2)已知等差数列 an 的前n项和为Sn，且满足S33S221，则数列 an 的公差是()B 1 C 2 D 3答案(1)A(2)C解析(1)an为等差数列，a1a52a3，a1a3a53a33，得a31，S55a1a525a35.故选 A.(2)Snna1an2，Snna1an2，又S33S221，得a1

14、a32a1a221，即a3a22，文档编码:CZ7G2C7D10P8 HE5I10G8M6L9 ZU10E6P8I3I2文档编码:CZ7G2C7D10P8 HE5I10G8M6L9 ZU10E6P8I3I2文档编码:CZ7G2C7D10P8 HE5I10G8M6L9 ZU10E6P8I3I2文档编码:CZ7G2C7D10P8 HE5I10G8M6L9 ZU10E6P8I3I2文档编码:CZ7G2C7D10P8 HE5I10G8M6L9 ZU10E6P8I3I2文档编码:CZ7G2C7D10P8 HE5I10G8M6L9 ZU10E6P8I3I2文档编码:CZ7G2C7D10P8 HE5I10G

15、8M6L9 ZU10E6P8I3I2文档编码:CZ7G2C7D10P8 HE5I10G8M6L9 ZU10E6P8I3I2文档编码:CZ7G2C7D10P8 HE5I10G8M6L9 ZU10E6P8I3I2文档编码:CZ7G2C7D10P8 HE5I10G8M6L9 ZU10E6P8I3I2文档编码:CZ7G2C7D10P8 HE5I10G8M6L9 ZU10E6P8I3I2文档编码:CZ7G2C7D10P8 HE5I10G8M6L9 ZU10E6P8I3I2文档编码:CZ7G2C7D10P8 HE5I10G8M6L9 ZU10E6P8I3I2文档编码:CZ7G2C7D10P8 HE5I10

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18、10G8M6L9 ZU10E6P8I3I2文档编码:CZ7G2C7D10P8 HE5I10G8M6L9 ZU10E6P8I3I2文档编码:CZ7G2C7D10P8 HE5I10G8M6L9 ZU10E6P8I3I2文档编码:CZ7G2C7D10P8 HE5I10G8M6L9 ZU10E6P8I3I2文档编码:CZ7G2C7D10P8 HE5I10G8M6L9 ZU10E6P8I3I2文档编码:CZ7G2C7D10P8 HE5I10G8M6L9 ZU10E6P8I3I2文档编码:CZ7G2C7D10P8 HE5I10G8M6L9 ZU10E6P8I3I2文档编码:CZ7G2C7D10P8 HE5

19、I10G8M6L9 ZU10E6P8I3I2文档编码:CZ7G2C7D10P8 HE5I10G8M6L9 ZU10E6P8I3I2文档编码:CZ7G2C7D10P8 HE5I10G8M6L9 ZU10E6P8I3I2文档编码:CZ7G2C7D10P8 HE5I10G8M6L9 ZU10E6P8I3I2文档编码:CZ7G2C7D10P8 HE5I10G8M6L9 ZU10E6P8I3I2文档编码:CZ7G2C7D10P8 HE5I10G8M6L9 ZU10E6P8I3I2文档编码:CZ7G2C7D10P8 HE5I10G8M6L9 ZU10E6P8I3I2文档编码:CZ7G2C7D10P8 HE

20、5I10G8M6L9 ZU10E6P8I3I2文档编码:CZ7G2C7D10P8 HE5I10G8M6L9 ZU10E6P8I3I2文档编码:CZ7G2C7D10P8 HE5I10G8M6L9 ZU10E6P8I3I2文档编码:CZ7G2C7D10P8 HE5I10G8M6L9 ZU10E6P8I3I2文档编码:CZ7G2C7D10P8 HE5I10G8M6L9 ZU10E6P8I3I2文档编码:CZ7G2C7D10P8 HE5I10G8M6L9 ZU10E6P8I3I2文档编码:CZ7G2C7D10P8 HE5I10G8M6L9 ZU10E6P8I3I2数列 an的公差为2.题型二等差数列的

21、判定与证明例 2 已知数列 an 中，a135，an21an1(n2，nN*)，数列 bn 满足bn1an 1(nN*)(1)求证：数列 bn 是等差数列；(2)求数列 an中的最大项和最小项，并说明理由(1)证明因为an21an1(n2，nN*)，bn1an1(nN*)，所以bn 1bn1an 1 11an1121an11an1anan11an11.又b11a1 152.所以数列 bn 是以52为首项，1 为公差的等差数列(2)解由(1)知bnn72，则an11bn122n7.设f(x)122x7，则f(x)在区间(，72)和(72，)上为减函数所以当n3 时，an取得最小值1，当n4 时，

22、an取得最大值3.引申探究例 2 中，若条件变为a135，nan1(n1)ann(n1)，探求数列 an的通项公式解由已知可得an1n1ann1，即an 1n1ann 1，又a135，文档编码:CZ7G2C7D10P8 HE5I10G8M6L9 ZU10E6P8I3I2文档编码:CZ7G2C7D10P8 HE5I10G8M6L9 ZU10E6P8I3I2文档编码:CZ7G2C7D10P8 HE5I10G8M6L9 ZU10E6P8I3I2文档编码:CZ7G2C7D10P8 HE5I10G8M6L9 ZU10E6P8I3I2文档编码:CZ7G2C7D10P8 HE5I10G8M6L9 ZU10E

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27、U10E6P8I3I2文档编码:CZ7G2C7D10P8 HE5I10G8M6L9 ZU10E6P8I3I2文档编码:CZ7G2C7D10P8 HE5I10G8M6L9 ZU10E6P8I3I2文档编码:CZ7G2C7D10P8 HE5I10G8M6L9 ZU10E6P8I3I2文档编码:CZ7G2C7D10P8 HE5I10G8M6L9 ZU10E6P8I3I2文档编码:CZ7G2C7D10P8 HE5I10G8M6L9 ZU10E6P8I3I2文档编码:CZ7G2C7D10P8 HE5I10G8M6L9 ZU10E6P8I3I2文档编码:CZ7G2C7D10P8 HE5I10G8M6L9

28、ZU10E6P8I3I2文档编码:CZ7G2C7D10P8 HE5I10G8M6L9 ZU10E6P8I3I2文档编码:CZ7G2C7D10P8 HE5I10G8M6L9 ZU10E6P8I3I2文档编码:CZ7G2C7D10P8 HE5I10G8M6L9 ZU10E6P8I3I2文档编码:CZ7G2C7D10P8 HE5I10G8M6L9 ZU10E6P8I3I2文档编码:CZ7G2C7D10P8 HE5I10G8M6L9 ZU10E6P8I3I2文档编码:CZ7G2C7D10P8 HE5I10G8M6L9 ZU10E6P8I3I2文档编码:CZ7G2C7D10P8 HE5I10G8M6L9

29、 ZU10E6P8I3I2文档编码:CZ7G2C7D10P8 HE5I10G8M6L9 ZU10E6P8I3I2ann是以a1135为首项，1为公差的等差数列，ann35(n1)1n25，ann225n.思维升华等差数列的四个判定方法(1)定义法：证明对任意正整数n都有an1an等于同一个常数(2)等差中项法：证明对任意正整数n都有 2an 1anan 2后，可递推得出an2an1an 1ananan1an1an2a2a1，根据定义得出数列an为等差数列(3)通项公式法：得出anpnq后，得an 1anp对任意正整数n恒成立，根据定义判定数列 an为等差数列(4)前n项和公式法：得出SnAn2

30、Bn后，根据Sn，an的关系，得出an，再使用定义法证明数列 an为等差数列(1)若an 是公差为1 的等差数列，则a2n1 2a2n 是()A公差为 3 的等差数列B公差为4 的等差数列C公差为 6 的等差数列D公差为9 的等差数列(2)在数列 an中，若a11，a212，2an11an1an2(nN*)，则该数列的通项为()Aan1nBan2n1Can2n2Dan3n答案(1)C(2)A解析(1)a2n1 2a2n(a2n32a2n2)(a2n1a2n3)2(a2na2n2)222 6，a2n12a2n是公差为6 的等差数列(2)由已知式2an11an1an 2可得1an11an1an21

31、an1，知1an 是首项为1a11，公差为1a21a1211 的等差数列，所以1ann，即an1n.文档编码:CZ7G2C7D10P8 HE5I10G8M6L9 ZU10E6P8I3I2文档编码:CZ7G2C7D10P8 HE5I10G8M6L9 ZU10E6P8I3I2文档编码:CZ7G2C7D10P8 HE5I10G8M6L9 ZU10E6P8I3I2文档编码:CZ7G2C7D10P8 HE5I10G8M6L9 ZU10E6P8I3I2文档编码:CZ7G2C7D10P8 HE5I10G8M6L9 ZU10E6P8I3I2文档编码:CZ7G2C7D10P8 HE5I10G8M6L9 ZU10

32、E6P8I3I2文档编码:CZ7G2C7D10P8 HE5I10G8M6L9 ZU10E6P8I3I2文档编码:CZ7G2C7D10P8 HE5I10G8M6L9 ZU10E6P8I3I2文档编码:CZ7G2C7D10P8 HE5I10G8M6L9 ZU10E6P8I3I2文档编码:CZ7G2C7D10P8 HE5I10G8M6L9 ZU10E6P8I3I2文档编码:CZ7G2C7D10P8 HE5I10G8M6L9 ZU10E6P8I3I2文档编码:CZ7G2C7D10P8 HE5I10G8M6L9 ZU10E6P8I3I2文档编码:CZ7G2C7D10P8 HE5I10G8M6L9 ZU1

33、0E6P8I3I2文档编码:CZ7G2C7D10P8 HE5I10G8M6L9 ZU10E6P8I3I2文档编码:CZ7G2C7D10P8 HE5I10G8M6L9 ZU10E6P8I3I2文档编码:CZ7G2C7D10P8 HE5I10G8M6L9 ZU10E6P8I3I2文档编码:CZ7G2C7D10P8 HE5I10G8M6L9 ZU10E6P8I3I2文档编码:CZ7G2C7D10P8 HE5I10G8M6L9 ZU10E6P8I3I2文档编码:CZ7G2C7D10P8 HE5I10G8M6L9 ZU10E6P8I3I2文档编码:CZ7G2C7D10P8 HE5I10G8M6L9 ZU

34、10E6P8I3I2文档编码:CZ7G2C7D10P8 HE5I10G8M6L9 ZU10E6P8I3I2文档编码:CZ7G2C7D10P8 HE5I10G8M6L9 ZU10E6P8I3I2文档编码:CZ7G2C7D10P8 HE5I10G8M6L9 ZU10E6P8I3I2文档编码:CZ7G2C7D10P8 HE5I10G8M6L9 ZU10E6P8I3I2文档编码:CZ7G2C7D10P8 HE5I10G8M6L9 ZU10E6P8I3I2文档编码:CZ7G2C7D10P8 HE5I10G8M6L9 ZU10E6P8I3I2文档编码:CZ7G2C7D10P8 HE5I10G8M6L9 Z

35、U10E6P8I3I2文档编码:CZ7G2C7D10P8 HE5I10G8M6L9 ZU10E6P8I3I2文档编码:CZ7G2C7D10P8 HE5I10G8M6L9 ZU10E6P8I3I2文档编码:CZ7G2C7D10P8 HE5I10G8M6L9 ZU10E6P8I3I2文档编码:CZ7G2C7D10P8 HE5I10G8M6L9 ZU10E6P8I3I2文档编码:CZ7G2C7D10P8 HE5I10G8M6L9 ZU10E6P8I3I2文档编码:CZ7G2C7D10P8 HE5I10G8M6L9 ZU10E6P8I3I2文档编码:CZ7G2C7D10P8 HE5I10G8M6L9

36、ZU10E6P8I3I2文档编码:CZ7G2C7D10P8 HE5I10G8M6L9 ZU10E6P8I3I2文档编码:CZ7G2C7D10P8 HE5I10G8M6L9 ZU10E6P8I3I2文档编码:CZ7G2C7D10P8 HE5I10G8M6L9 ZU10E6P8I3I2文档编码:CZ7G2C7D10P8 HE5I10G8M6L9 ZU10E6P8I3I2文档编码:CZ7G2C7D10P8 HE5I10G8M6L9 ZU10E6P8I3I2文档编码:CZ7G2C7D10P8 HE5I10G8M6L9 ZU10E6P8I3I2文档编码:CZ7G2C7D10P8 HE5I10G8M6L9

37、 ZU10E6P8I3I2文档编码:CZ7G2C7D10P8 HE5I10G8M6L9 ZU10E6P8I3I2文档编码:CZ7G2C7D10P8 HE5I10G8M6L9 ZU10E6P8I3I2文档编码:CZ7G2C7D10P8 HE5I10G8M6L9 ZU10E6P8I3I2文档编码:CZ7G2C7D10P8 HE5I10G8M6L9 ZU10E6P8I3I2文档编码:CZ7G2C7D10P8 HE5I10G8M6L9 ZU10E6P8I3I2文档编码:CZ7G2C7D10P8 HE5I10G8M6L9 ZU10E6P8I3I2文档编码:CZ7G2C7D10P8 HE5I10G8M6L

38、9 ZU10E6P8I3I2文档编码:CZ7G2C7D10P8 HE5I10G8M6L9 ZU10E6P8I3I2文档编码:CZ7G2C7D10P8 HE5I10G8M6L9 ZU10E6P8I3I2文档编码:CZ7G2C7D10P8 HE5I10G8M6L9 ZU10E6P8I3I2文档编码:CZ7G2C7D10P8 HE5I10G8M6L9 ZU10E6P8I3I2文档编码:CZ7G2C7D10P8 HE5I10G8M6L9 ZU10E6P8I3I2文档编码:CZ7G2C7D10P8 HE5I10G8M6L9 ZU10E6P8I3I2文档编码:CZ7G2C7D10P8 HE5I10G8M6

39、L9 ZU10E6P8I3I2文档编码:CZ7G2C7D10P8 HE5I10G8M6L9 ZU10E6P8I3I2文档编码:CZ7G2C7D10P8 HE5I10G8M6L9 ZU10E6P8I3I2文档编码:CZ7G2C7D10P8 HE5I10G8M6L9 ZU10E6P8I3I2文档编码:CZ7G2C7D10P8 HE5I10G8M6L9 ZU10E6P8I3I2文档编码:CZ7G2C7D10P8 HE5I10G8M6L9 ZU10E6P8I3I2文档编码:CZ7G2C7D10P8 HE5I10G8M6L9 ZU10E6P8I3I2文档编码:CZ7G2C7D10P8 HE5I10G8M

40、6L9 ZU10E6P8I3I2文档编码:CZ7G2C7D10P8 HE5I10G8M6L9 ZU10E6P8I3I2文档编码:CZ7G2C7D10P8 HE5I10G8M6L9 ZU10E6P8I3I2文档编码:CZ7G2C7D10P8 HE5I10G8M6L9 ZU10E6P8I3I2文档编码:CZ7G2C7D10P8 HE5I10G8M6L9 ZU10E6P8I3I2文档编码:CZ7G2C7D10P8 HE5I10G8M6L9 ZU10E6P8I3I2文档编码:CZ7G2C7D10P8 HE5I10G8M6L9 ZU10E6P8I3I2文档编码:CZ7G2C7D10P8 HE5I10G8

41、M6L9 ZU10E6P8I3I2文档编码:CZ7G2C7D10P8 HE5I10G8M6L9 ZU10E6P8I3I2文档编码:CZ7G2C7D10P8 HE5I10G8M6L9 ZU10E6P8I3I2文档编码:CZ7G2C7D10P8 HE5I10G8M6L9 ZU10E6P8I3I2文档编码:CZ7G2C7D10P8 HE5I10G8M6L9 ZU10E6P8I3I2文档编码:CZ7G2C7D10P8 HE5I10G8M6L9 ZU10E6P8I3I2文档编码:CZ7G2C7D10P8 HE5I10G8M6L9 ZU10E6P8I3I2文档编码:CZ7G2C7D10P8 HE5I10G

42、8M6L9 ZU10E6P8I3I2文档编码:CZ7G2C7D10P8 HE5I10G8M6L9 ZU10E6P8I3I2文档编码:CZ7G2C7D10P8 HE5I10G8M6L9 ZU10E6P8I3I2文档编码:CZ7G2C7D10P8 HE5I10G8M6L9 ZU10E6P8I3I2文档编码:CZ7G2C7D10P8 HE5I10G8M6L9 ZU10E6P8I3I2文档编码:CZ7G2C7D10P8 HE5I10G8M6L9 ZU10E6P8I3I2文档编码:CZ7G2C7D10P8 HE5I10G8M6L9 ZU10E6P8I3I2文档编码:CZ7G2C7D10P8 HE5I10

43、G8M6L9 ZU10E6P8I3I2文档编码:CZ7G2C7D10P8 HE5I10G8M6L9 ZU10E6P8I3I2文档编码:CZ7G2C7D10P8 HE5I10G8M6L9 ZU10E6P8I3I2文档编码:CZ7G2C7D10P8 HE5I10G8M6L9 ZU10E6P8I3I2文档编码:CZ7G2C7D10P8 HE5I10G8M6L9 ZU10E6P8I3I2文档编码:CZ7G2C7D10P8 HE5I10G8M6L9 ZU10E6P8I3I2文档编码:CZ7G2C7D10P8 HE5I10G8M6L9 ZU10E6P8I3I2文档编码:CZ7G2C7D10P8 HE5I1

44、0G8M6L9 ZU10E6P8I3I2文档编码:CZ7G2C7D10P8 HE5I10G8M6L9 ZU10E6P8I3I2文档编码:CZ7G2C7D10P8 HE5I10G8M6L9 ZU10E6P8I3I2文档编码:CZ7G2C7D10P8 HE5I10G8M6L9 ZU10E6P8I3I2文档编码:CZ7G2C7D10P8 HE5I10G8M6L9 ZU10E6P8I3I2文档编码:CZ7G2C7D10P8 HE5I10G8M6L9 ZU10E6P8I3I2文档编码:CZ7G2C7D10P8 HE5I10G8M6L9 ZU10E6P8I3I2题型三等差数列的性质及应用命题点 1 等差数

45、列的性质例 3(1)(2015广东)在等差数列 an 中，若a3a4a5a6a725，则a2a8_.(2)已知等差数列 an 的前n项和为Sn，且S10 10，S2030，则S30_.答案(1)10(2)60解析(1)因为 an 是等差数列，所以a3a7a4a6a2a8 2a5，a3a4a5a6a75a525，即a55，a2a82a510.(2)S10，S20S10，S30S20成等差数列，且S1010，S2030，S20S1020，S30 3010210 30，S3060.命题点 2 等差数列前n项和的最值例 4 在等差数列 an 中，已知a120，前n项和为Sn，且S10S15，求当n取何

46、值时，Sn取得最大值，并求出它的最大值解a120，S10S15，10201092d152015142d，d53.方法一由an20(n1)5353n653.得a130.即当n12 时，an0，当n14 时，an0.当n12 或 13 时，Sn取得最大值，且最大值为S12S13122012112 53130.方法二Sn 20nnn12 5356n21256n56n25223 12524.nN*，当n12 或 13 时，Sn有最大值，且最大值为S12S13130.方法三由S10S15得a11a12a13a14a150.5a130，即a13 0.文档编码:CZ7G2C7D10P8 HE5I10G8M6

47、L9 ZU10E6P8I3I2文档编码:CZ7G2C7D10P8 HE5I10G8M6L9 ZU10E6P8I3I2文档编码:CZ7G2C7D10P8 HE5I10G8M6L9 ZU10E6P8I3I2文档编码:CZ7G2C7D10P8 HE5I10G8M6L9 ZU10E6P8I3I2文档编码:CZ7G2C7D10P8 HE5I10G8M6L9 ZU10E6P8I3I2文档编码:CZ7G2C7D10P8 HE5I10G8M6L9 ZU10E6P8I3I2文档编码:CZ7G2C7D10P8 HE5I10G8M6L9 ZU10E6P8I3I2文档编码:CZ7G2C7D10P8 HE5I10G8M

48、6L9 ZU10E6P8I3I2文档编码:CZ7G2C7D10P8 HE5I10G8M6L9 ZU10E6P8I3I2文档编码:CZ7G2C7D10P8 HE5I10G8M6L9 ZU10E6P8I3I2文档编码:CZ7G2C7D10P8 HE5I10G8M6L9 ZU10E6P8I3I2文档编码:CZ7G2C7D10P8 HE5I10G8M6L9 ZU10E6P8I3I2文档编码:CZ7G2C7D10P8 HE5I10G8M6L9 ZU10E6P8I3I2文档编码:CZ7G2C7D10P8 HE5I10G8M6L9 ZU10E6P8I3I2文档编码:CZ7G2C7D10P8 HE5I10G8

49、M6L9 ZU10E6P8I3I2文档编码:CZ7G2C7D10P8 HE5I10G8M6L9 ZU10E6P8I3I2文档编码:CZ7G2C7D10P8 HE5I10G8M6L9 ZU10E6P8I3I2文档编码:CZ7G2C7D10P8 HE5I10G8M6L9 ZU10E6P8I3I2文档编码:CZ7G2C7D10P8 HE5I10G8M6L9 ZU10E6P8I3I2文档编码:CZ7G2C7D10P8 HE5I10G8M6L9 ZU10E6P8I3I2文档编码:CZ7G2C7D10P8 HE5I10G8M6L9 ZU10E6P8I3I2文档编码:CZ7G2C7D10P8 HE5I10G

50、8M6L9 ZU10E6P8I3I2文档编码:CZ7G2C7D10P8 HE5I10G8M6L9 ZU10E6P8I3I2文档编码:CZ7G2C7D10P8 HE5I10G8M6L9 ZU10E6P8I3I2文档编码:CZ7G2C7D10P8 HE5I10G8M6L9 ZU10E6P8I3I2文档编码:CZ7G2C7D10P8 HE5I10G8M6L9 ZU10E6P8I3I2文档编码:CZ7G2C7D10P8 HE5I10G8M6L9 ZU10E6P8I3I2文档编码:CZ7G2C7D10P8 HE5I10G8M6L9 ZU10E6P8I3I2文档编码:CZ7G2C7D10P8 HE5I10