高一数学等差数列的性质 人教版 课件.ppt

单击此处编辑母版标题样式,单击此处编辑母版文本样式,第二级,第三级,第四级,第五级,*,等差数列的性质,知识回顾,等差数列,AAAAAAAAAAAAA,每一项与,它前一项的,差,如果一个数列从第,2,项起，,等于同一个常数,.,.,.,【,说明,】,AAA,数列,a,n,为等差数列,a,n,+1,-,a,n,=d,或,a,n,+1,=,a,n,+d,d,=a,n,+1,-,a,n,公差是,唯一,的常数。,a,n,=a,1,+,(,n-,1),d,等差数列各项对应的点都在同一条直线上,.,一、判定题：下列数列是否是等差数列？,.,9，7，5，3，,-2n+11，；,.,-,1，11，23，35，12n-13，；,.,1，2，1，2，；,.,1，2，4，6，8，10，；,.,a,a,a,a,，,a，,；,复习巩固：,(1),等差数列,8,，,5,，,2,，,，的第,5,项是,AA,AAAAAAA,(2),等差数列,-5,，,-9,，,-13,，,的第,n,项是,A,-4,a,n,=-5+(n-1),.,(-4),10,【,说明,】,在等差数列,a,n,的通项公式中,a,1,、,d,、,a,n,、,n,任知,三,个，即,可求出,另外一个,二、填空题：,(3),已知,a,n,为等差数列，,a,1,=3,，,d,=2,，,a,n,=21,，则,n,=,【,说明,】,3.,更一般的情形，,an,=,，,d,=,等差数列的性质,1.,a,n,为等差数列,2.,a,、,b,、,c,成等差数列,a,n,+1,-,a,n,=d,a,n,+1,=a,n,+d,a,n,=,a,1,+,(,n-,1),d,a,m,+,(,n,-,m,),d,b,为,a,、,c,的等差中项,AA,2,b=a+c,4.,在,等差数列,a,n,中，由,m+n=p+q,a,m,+,a,n,=,a,p,+,a,q,例,1.,梯子的最高一级宽,33 cm,，,最低一级宽,110 cm,，,中间,还有,10,级，各级的宽度成等差数列，计算中间各级的宽,.,项数（上）,1,2,3,4,5,6,7,8,9,10,11,12,数列的项,33,110,项数（下）,12,11,10,9,8,7,6,5,4,3,2,1,分析：,例题分析,解法一,:,用,a,n,表示题中的等差数列，由已知条件，有,a,1,=33,，,a,12,=110,，,n=12,又,a,12,=a,1,+(121)d,即,110,33,11d,所以,d=7,因此，,a,2,=33+7=40 a,3,=40+47 a,11,=96+7=103,答：梯子中间各级的宽从上到下依次是,40cm,、,47cm,、,54cm,、,61cm,、,68m,、,75cm,、,82cm,、,89cm,、,96cm,、,103cm.,例,2.,在,等差数列,a,n,中,(1),已知,a,6,+,a,9,+,a,12,+,a,15,=20,，求,a,1,+,a,20,例题分析,(2,）,已知,a,3,+,a,11,=10,，,求,a,6,+,a,7,+,a,8,(3),已知,a,4,+,a,5,+,a,6,+,a,7,=56,，,a,4,a,7,=187,，求,a,14,及公差,d,.,分析：由,a,1,+,a,20,=,a,6,+,a,15,=,a,9,+,a,12,及,a,6,+,a,9,+,a,12,+,a,15,=20,，,可得,a,1,+,a,20,=10,分析：,a,3,+,a,11,=,a,6,+,a,8,=2,a,7,，,又已知,a,3,+,a,11,=10,，,a,6,+,a,7,+,a,8,=,（,a,3,+,a,11,）,=15,分析：,a,4,+,a,5,+,a,6,+,a,7,=56,a,4,+,a,7,=28 ,又,a,4,a,7,=187,，,解,、,得,a,4,=17,a,7,=11,a,4,=11,a,7,=17,或,d=,_,2,或,2,从而,a,14,=,_,3,或,31,课堂练习,1,.,等差数列,a,n,的前三项依次为,a,-6,，,2,a,-5,，,-3,a+,2,，,则,a,等于（,),A,.-,1,B,.,1,C,.,-2,D.2,B,2,.,在,数列,a,n,中,a,1,=1,，,a,n,=,a,n+1,+4,，则,a,10,=,2(2,a,-5)=(-3,a+2,)+(,a,-6,),提示,1:,提示：,d=a,n+,1,a,n,=4,-35,3.,在,等差数列,a,n,中,(1),若,a,59,=70,，,a,80,=112,，求,a,101,；,(2),若,a,p,=,q,，,a,q,=,p,(,pq,),，求,a,p+q,d=,2,a,101,=154,d=,-1,a,p+q,=,0,研究性问题,300 500,4,.,在,等差数列,a,n,中,a,1,=83,，,a,4,=98,，,则这个数列有,多少项在,300,到,500,之间？,d=,5,提示：,a,n,=78+5,n,n45，46，84,40,2.,已知,a,n,为等差数列，若,a,10,=20,，,d,=-1,，求,a,3,?,1.若,a,12,=23，,a,42,=143，,a,n,=263，求,n,.,3.,三数成等差数列，它们的和为,12,，首尾二数的,积为,12,，求此三数,.,d=,4,n,=72,a,3,=,a,10,+(3-10)d,a,3,=27,设这三个数分别为,a-d a,，,a+d,，则,3a=12,，,a,2,-d,2,=12,6，4，2或2，4，6,【,说明,】,3.,更一般的情形，,a,n,=,，,d,=,一、知识巩固,1.,a,n,为等差数列,2.,a,、,b,、,c,成等差数列,a,n,+1,-,a,n,=d,a,n,+1,=a,n,+d,a,n,=,a,1,+,(,n-,1),d,a,m,+,(,n,-,m,),d,b,为,a,、,c,的等差中项,AA,2,b=a+c,4.,在,等差数列,a,n,中，由,m+n=p+q,a,m,+,a,n,=,a,p,+,a,q,5,.,在等差数列,a,n,中,a,1,+,a,n,a,2,+,a,n-,1,a,3,+,a,n-,2,=,=,=,前,100,个自然数的和：,1+2+3+100=,；,前,n,个奇数的和：,1+3+5+(2,n,-1)=,；,前,n,个偶数的和：,2+4+6+2,n,=,.,思考题：,如何求下列和n,2,n,(,n,+1),二、学习新课,等差数列前,n,项和,S,n,=,=,.,S,n,=,a,1,+,a,2,+,a,3,+,a,n-,2,+,a,n-,1,+,a,n,(1),S,n,=,a,n,+,a,n-,1,+,a,n-,2,+,a,3,+,a,2,+,a,1,(2),(1)+(2)得,2S,n,=,n,(,a,1,+,a,n,),【,说明,】,推导等差数列的前,n,项和公式的方法叫,；,等差数列的前,n,项和公式类同于,；,倒序相加法,梯形的面积公式,等差数列,a,n,的首项为,a,1,，,公差为,d,，,项数为,n,，第,n,项为,a,n,，前,n,项和为,S,n,，,请填写下表：,三、课堂练习,a,1,d,n,a,n,s,n,5,10,10,-2,50,2550,-38,-10,-360,14.5,26,32,95,500,100,2,2,15,0.7,604.5,例,2,如图，一个堆放铅笔的,V,形架的最下面一层放,1,支,铅笔，往上每一层都比它下面一层多放一支，最,上面一层放,120,支,.,这个,V,形架上共放着多少支铅笔？,