高考数学复习第六章数列6.2等差数列及其前n项和市赛课公开课一等奖省名师优质课获奖课件.pptx

剖析题型 提炼方法,实验解读,构建知识网络 强化答题语句,探究高考 明确考向,6.2,等差数列及其前,n,项和,第六章数列,1/75,基础知识自主学习,课时作业,题型分类深度剖析,内容索引,2/75,基础知识自主学习,3/75,1.,等差数列定义,普通地，假如一个数列,_,，那么这个数列就叫做等差数列，这个常数叫做等差数列,，通惯用字母,表示,.,2.,等差数列通项公式,假如等差数列,a,n,首项为,a,1,，公差为,d,，那么它通项公式是,.,知识梳理,从第,2,项起，每一项与它前一项差等于同一个,常数,公差,d,a,n,a,1,(,n,1),d,4/75,3.,等差中项,由三个数,a,，,A,，,b,组成等差数列能够看成最简单等差数列,.,这时，,A,叫做,a,与,b,.,等差中项,5/75,4.,等差数列惯用性质,(1),通项公式推广：,a,n,a,m,(,n,，,m,N,*,).,(2),若,a,n,为等差数列，且,k,l,m,n,(,k,，,l,，,m,，,n,N,*,),，则,.,(3),若,a,n,是等差数列，公差为,d,，则,a,2,n,也是等差数列，公差为,.,(4),若,a,n,，,b,n,是等差数列，则,pa,n,qb,n,也是等差数列,.,(5),若,a,n,是等差数列，公差为,d,，则,a,k,，,a,k,m,，,a,k,2,m,，,(,k,，,m,N,*,),是公差为,等差数列,.,(6),数列,S,m,，,S,2,m,S,m,，,S,3,m,S,2,m,，,组成等差数列,.,(,n,m,),d,a,k,a,l,a,m,a,n,2,d,md,6/75,5.,等差数列前,n,项和公式,6.,等差数列前,n,项和公式与函数关系,数列,a,n,是等差数列,S,n,An,2,Bn,(,A,，,B,为常数,).,7.,等差数列前,n,项和最值,在等差数列,a,n,中，,a,1,0,，,d,0,，则,S,n,存在最,值；若,a,1,0,，则,S,n,存在最,值,.,大,小,设等差数列,a,n,公差为,d,，其前,n,项和,S,n,_,或,S,n,_,.,7/75,等差数列四种判断方法,(1),定义法：,a,n,1,a,n,d,(,d,是常数,),a,n,是等差数列,.,(2),等差中项法：,2,a,n,1,a,n,a,n,2,(,n,N,*,),a,n,是等差数列,.,(3),通项公式：,a,n,pn,q,(,p,，,q,为常数,),a,n,是等差数列,.,(4),前,n,项和公式：,S,n,An,2,Bn,(,A,，,B,为常数,),a,n,是等差数列,.,【,知识拓展,】,8/75,题组一思索辨析,1.,判断以下结论是否正确,(,请在括号中打,“”,或,“”,),(1),若一个数列从第二项起每一项与它前一项差都是常数，则这个数列是等差数列,.(,),(2),等差数列,a,n,单调性是由公差,d,决定,.(,),(3),等差数列前,n,项和公式是常数项为,0,二次函数,.(,),基础自测,1,2,3,4,5,6,9/75,(4),已知等差数列,a,n,通项公式,a,n,3,2,n,，则它公差为,2.(,),(5),数列,a,n,为等差数列充要条件是对任意,n,N,*,，都有,2,a,n,1,a,n,a,n,2,.,(,),(6),已知数列,a,n,通项公式是,a,n,pn,q,(,其中,p,，,q,为常数,),，则数列,a,n,一定是等差数列,.(,),1,2,3,4,5,6,10/75,题组二教材改编,2.P46A,组,T2,设数列,a,n,是等差数列，其前,n,项和为,S,n,，若,a,6,2,且,S,5,30,，则,S,8,等于,A.31 B.32,C.33 D.34,答案,解析,1,2,3,4,5,6,11/75,1,2,3,4,5,6,12/75,3.P39T5,在等差数列,a,n,中，若,a,3,a,4,a,5,a,6,a,7,450,，则,a,2,a,8,_.,答案,1,2,3,4,5,6,180,解析,由等差数列性质，得,a,3,a,4,a,5,a,6,a,7,5,a,5,450,，,a,5,90,，,a,2,a,8,2,a,5,180.,解析,13/75,题组三易错自纠,4.,一个等差数列首项为,，从第,10,项起开始比,1,大，则这个等差数列公差,d,取值范围是,答案,1,2,3,4,5,6,解析,14/75,1,2,3,4,5,6,15/75,5.,若等差数列,a,n,满足,a,7,a,8,a,9,0,，,a,7,a,10,0,，,a,2 016,a,2 017,0,，,a,2 016,a,2 017,0,成立最大正整数,n,是,A.2 016 B.2 017,C.4 032 D.4 033,解析,答案,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,53/75,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,解析,因为,a,1,0,，,a,2 016,a,2 017,0,，,a,2 016,a,2 017,0,，,所以,d,0,，,a,2 017,0,成立最大正整数,n,是,4 032,，故选,C.,54/75,解析,答案,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,7.(,安徽省安师大附中、马鞍山二中阶段性测试,),若等差数列,a,n,前,n,项和为,S,n,，且满足,a,2,S,3,4,，,a,3,S,5,12,，则,a,4,S,7,值是,_.,24,55/75,解析,因为,a,4,和,a,2 016,是,3,x,2,12,x,4,0,两根，所以,a,4,a,2 016,4.,又,a,4,，,a,1 010,，,a,2 016,成等差数列，所以,2,a,1 010,a,4,a,2 016,，即,a,1 010,2,，,所以,8.,等差数列,a,n,中,a,4,，,a,2 016,是,3,x,2,12,x,4,0,两根，则,a,1 010,_.,解析,答案,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,56/75,解析,答案,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,9.(,郑州模拟,),张丘建算经卷上第,22,题为：,“,今有女善织，日益功疾,.,初日织五尺，今一月日织九匹三丈,.,”,其意思为今有女子善织布，且从第,2,天起，天天比前一天多织相同量布,.,若第一天织,5,尺布，现在一个月,(,按,30,天计,),共织,390,尺布，则该女最终一天织,_,尺布,.,21,57/75,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,58/75,10.,设数列,a,n,通项公式为,a,n,2,n,10(,n,N,*,),，则,|,a,1,|,|,a,2,|,|,a,15,|,_.,解析,答案,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,130,解析,由,a,n,2,n,10(,n,N,*,),知,a,n,是以,8,为首项，,2,为公差等差数列，,又由,a,n,2,n,10,0,，得,n,5,，,当,n,5,时，,a,n,0,，当,n,5,时，,a,n,0,，,|,a,1,|,|,a,2,|,|,a,15,|,(,a,1,a,2,a,3,a,4,),(,a,5,a,6,a,15,),20,110,130.,59/75,解答,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,11.(,全国,),等差数列,a,n,中，,a,3,a,4,4,，,a,5,a,7,6.,(1),求,a,n,通项公式；,60/75,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,解,设数列,a,n,首项为,a,1,，公差为,d,，,61/75,解答,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,(2),设,b,n,a,n,，求数列,b,n,前,10,项和，其中,x,表示不超出,x,最大整数，如,0.9,0,，,2.6,2.,62/75,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,所以数列,b,n,前,10,项和为,1,3,2,2,3,3,4,2,24.,63/75,12.(,贵州质检,),已知数列,a,n,各项均为正数，前,n,项和为,S,n,，且满足,2,S,n,n,4(,n,N,*,).,(1),求证：数列,a,n,为等差数列；,证实,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,64/75,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,解得,a,1,3(,a,1,1,舍去,).,65/75,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,所以,a,n,1,a,n,1,或,a,n,1,a,n,1,.,若,a,n,1,a,n,1,，则,a,n,a,n,1,1.,而,a,1,3,，,所以,a,2,2,，这与数列,a,n,各项均为正数相矛盾，,所以,a,n,1,a,n,1,，即,a,n,a,n,1,1,，,所以数列,a,n,是首项为,3,，公差为,1,等差数列,.,66/75,解答,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,(2),求数列,a,n,通项公式,.,解,由,(1),知,a,1,3,，,d,1,，,所以数列,a,n,通项公式,a,n,3,(,n,1),1,n,2,，,即,a,n,n,2.,67/75,13.(,郑州一模,),设数列,a,n,满足：,a,1,1,，,a,2,3,，且,2,na,n,(,n,1),a,n,1,(,n,1),a,n,1,，则,a,20,值是,_.,技能提升练,解析,答案,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,68/75,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,解析,2,na,n,(,n,1),a,n,1,(,n,1),a,n,1,，,数列,na,n,是以,a,1,1,为首项，,2,a,2,a,1,5,为公差等差数列，,20,a,20,1,5,19,96,，,69/75,答案,解析,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,70/75,15.,设等差数列,a,n,前,n,项和为,S,n,，若,S,m,1,2,，,S,m,0,，,S,m,1,3,，则,m,_.,拓展冲刺练,答案,解析,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,5,71/75,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,解析,数列,a,n,为等差数列，且前,n,项和为,S,n,，,解得,m,5,，经检验符合题意,.,72/75,答案,解析,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,121,73/75,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,解析,设数列,a,n,公差为,d,，,化简可得,d,2,a,1,2,，,所以,a,n,1,(,n,1),2,2,n,1,，,74/75,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,75/75,