高一数学备课组.pptx

高一数学备课组,等差数列概念,推导等差数列通项公式的方法叫做,法,.,递推,每一项与,它前一项的差,等差数列,如果一个数列从第,2,项起，,等于同一个常数,.,.,.,【,说明,】,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,的常数；,唯一,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,.,知识回顾,由定义归纳通项公式,a,2,a,1,=d,，,a,3,a,2,=d,，,a,4,a,3,=d,，,则,a,2,=a,1,+d,a,3,=a,2,+d=a,1,+2d,a,4,=a,3,+d=a,1,+3d,由此得到,a,n,=a,1,+(n,1)d,a,n,1,a,n,2,=d,a,n,a,n,1,=d.,这（,n,1,）个式子迭加,a,n,a,1,=(n,1)d,当,n=1,时，上式两边均等于,a,1,，即等式也成立的。这表明当,nN,*,时上式都成立，因而它就是等差数列,a,n,的通项公式。,等差中项,如果在,a,与,b,中间插入一个数,A,，使,a,，,A,，,b,成等差数列，那么,A,叫做,a,与,b,的,等差中项,。,课本,P35.2.,3,，,4,，,5,，,课堂练习,结论归纳,:,数列,a,n,是公差为,d,的等差数列。,数列,a,1,a,3,a,5,a,7,是公差为,等差数列,数列,a,2,a,4,a,6,a,8,是公差为,等差数列,数列,ma,2,ma,4,ma,6,ma,8,是公差,为,等差数列,例,1,(1),已知数列,a,n,的通项公式是,a,n,=3,n,-1,，,求证：,a,n,为等差数列；,(2),已知数列,a,n,是等差数列，,求证：数列,a,n,+a,n+,1,也是等差数列,.,【,小结,】,数列,a,n,为等差数,；,证明一个数列为等差数列的方法是,：,.,a,n,=,kn+b,k,、,b,是常数,.,证明,：,a,n,+1,a,n,为一个常数,.,例题分析,例,2,(1),等差数列,11,，,8,，,5,，,，的第,19,项是,；,(2),等差数列,-5,，,-9,，,-13,，,的第,项是,-307,；,(3),已知,a,n,为等差数列，若,a,1,=3,，,d,=,，,a,n,=21,，,则,n,=,；,(4),已知,a,n,为等差数列，若,a,17,=,，,d,=,，则,a,10,=,.,-49,99,13,【,说明,】,在等差数列,a,n,的通项公式中,a,1,、,d,、,a,n,、,n,任知,个，可求,.,三,另外一个,上面的命题中的等式两边有,相 同 数 目,的项，如,a,1,+,a,2,=,a,3,成立吗？,【,说明,】,3.,更一般的情形，,a,n,=,，,d,=,等差数列的性质,1,1.,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,n,=,kn +b,（,k,、,b,为常数）,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,3.,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,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,）,=15,三数成等差数列，它们的和为,12,，首尾二数的,积为,12,，求此三数,.,已知,a,n,为等差数列,且,a,4,+,a,5,+,a,6,+,a,7,=56,，,a,4,a,7,=187,，求公差,d,.,