1.2.1.2等差数列的性质及应用

,-,23,-,第,2,课时,等差数列的性质及应用,第,2,课时,等差数列的性质及应用,1,.,体会等差数列与一次函数的关系,能够运用一次函数的性质解决等差数列问题,.,2,.,掌握等差中项的定义,能够运用定义解决有关问题,.,3,.,掌握等差数列性质的应用及实际应用,.,1,.,等差中项,如果三个数,a,A,b,成等差数列,那么,A,叫作,a,与,b,的等差中项,.,等差中项的性质,:,(1),A,是,a,与,b,的等差中项,则,(2),当,2,A=a+b,时,A,是,a,与,b,的等差中项,.,(3),如果三个数成等差数列,那么通常设这三个数为,a-d,a,a+d,这样可以在解题过程中减少运算量,.,(4),如果数列,a,n,满足,2,a,n,=a,n-,1,+a,n+,1,(,n,2,n,N,+,),那么数列,a,n,是等差数列,.,答案,:,A,2,.,等差数列的性质,若数列,a,n,是公差为,d,的等差数列,则,(1),当,d=,0,时,数列为常数列,;,当,d,0,时,数列为递增数列,;,当,d,0,时,数列为递减数列,.,(6),若数列,a,n,是有穷等差数列,则与首末两项等距离的两项之和都相等,且等于首末两项之和,即,a,1,+a,n,=a,2,+a,n-,1,=,=a,i+,1,+a,n-i,=,(,n,i,N,+,),.,(7),数列,a,n,+b,(,b,是常数,),是公差为,d,的等差数列,.,(8),下标成等差数列且公差为,m,的项,a,k,a,k+m,a,k+,2,m,(,k,m,N,+,),组成公差为,md,的等差数列,.,(9),若数列,b,n,也为等差数列,则,ka,n,+mb,n,+b,(,k,m,b,为常数,),也是等差数列,.,【做一做,2,】,在等差数列,a,n,中,a,3,+a,7,=,37,则,a,2,+a,4,+a,6,+a,8,=,.,解析,:,由等差数列的性质知,a,2,+a,4,+a,6,+a,8,=,2(,a,3,+a,7,),=,2,37,=,74,.,答案,:,74,题型一,题型二,题型三,题型一,等差中项的,应用,题型一,题型二,题型三,题型一,题型二,题型三,反思,证明三个数,(,式子,),成等差数列,一般可根据定义或等差中项将问题转化为证明等式成立,.,根据等差数列各项乘,(,或除以,),同一个常数,(,非零整数,),或加,(,或减,),同一个常数所得数列仍是等差数列,再结合问题条件亦可证明,.,题型一,题型二,题型三,【变式训练,1,】,在,-,1,与,7,之间顺次插入三个数,a,b,c,使这五个数成等差数列,求这个数列,.,解,:,-,1,a,b,c,7,成等差数列,b,是,-,1,与,7,的等差中项,a,是,-,1,与,b,的等差中项,c,是,b,与,7,的等差中项,题型一,题型二,题型三,题型二,等差数列性质的应用,【例,2,】,(1),已知,a,n,是等差数列,且,a,1,-a,4,+a,8,-a,12,+a,15,=,2,求,a,3,+a,13,的值,;,(2),在等差数列,a,n,中,若,a,49,=,80,a,59,=,100,求,a,79,.,分析,本题,(1),考查等差数列的性质,“,若,m+n=p+q,则,a,m,+a,n,=a,p,+a,q,”,的应用,;(2),考查性质,“,a,m,=a,n,+,(,m-n,),d,”,的应用,.,解,:,(1),a,n,是等差数列,a,1,+a,15,=a,4,+a,12,=a,3,+a,13,=,2,a,8,.,a,1,-a,4,+a,8,-a,12,+a,15,=,2,a,8,=,2,a,3,+a,13,=,2,a,8,=,2,2,=,4,.,(2),a,n,是等差数列,可设公差为,d.,由,a,59,=a,49,+,10,d,知,10,d=,100,-,80,解得,d=,2,.,a,79,=a,59,+,20,d,a,79,=,100,+,20,2,=,140,.,题型一,题型二,题型三,反思,在等差数列中,若,m+n=p+q=,2,k,则,a,m,+a,n,=a,p,+a,q,=,2,a,k,(,m,n,p,q,k,都是正整数,),它是一条重要性质,利用该性质可简化运算,.,题型一,题型二,题型三,【变式训练,2,】,已知等差数列,a,n,(1),若,a,2,+a,3,+a,25,+a,26,=,48,求,a,14,;,(2),若,a,2,+a,3,+a,4,+a,5,=,34,a,2,a,5,=,52,求公差,d.,解,:,(1),a,2,+a,26,=a,3,+a,25,=,2,a,14,a,2,+a,3,+a,25,+a,26,=,4,a,14,=,48,解得,a,14,=,12,.,题型一,题型二,题型三,(2),a,2,+a,5,=a,3,+a,4,a,2,+a,3,+a,4,+a,5,=,2(,a,2,+a,5,),=,34,.,即,a,2,+a,5,=,17,.,又已知,a,2,a,5,=,52,联立解得,a,2,=,4,a,5,=,13,或,a,2,=,13,a,5,=,4,.,题型一,题型二,题型三,题型三,实际应用问题,【例,3,】,梯子的最高一级宽,33 cm,最低一级宽,110 cm,中间还有,10,级,各级宽度依次成等差数列,计算中间各级的宽度,.,分析,:,要求梯子中间各级的宽度,必须知道各级宽度组成的等差数列的公差,.,又梯子的级数是,12,因此,问题相当于已知等差数列的首项、末项及项数求公差,.,解,:,设梯子的第,n,(1,n,12),级的宽为,a,n,cm,其中最高一级宽为,a,1,cm,则数列,a,n,是等差数列,.,由题意,得,a,1,=,33,a,12,=,110,n=,12,则,a,12,=a,1,+,11,d,所以,110,=,33,+,11,d,解得,d=,7,.,所以,a,2,=,33,+,7,=,40,a,3,=,40,+,7,=,47,a,11,=,96,+,7,=,103,即梯子中间各级的宽度从上到下依次是,40,cm,47,cm,54,cm,61,cm,68,cm,75,cm,82,cm,89,cm,96,cm,103,cm.,题型一,题型二,题型三,反思,解决实际应用问题的关键是建立数学模型,本题中的数学模型是已知等差数列的首项、末项及项数,求各项,.,题型一,题型二,题型三,【变式训练,3,】,甲虫是行动较快的昆虫之一,下表记录了某种类型的甲虫的爬行速度,:,(1),你能建立一个模型,表示甲虫的爬行距离和时间之间的关系吗,?,(2),利用建立的模型计算,甲虫,1 min,能爬多远,?,它爬行,49 cm,需要多长时间,?,题型一,题型二,题型三,解,:,(1),由表可知该数列从第,2,项起,每一项与前一项的差都是常数,9,.,8,所以是等差数列模型,.,因为,a,1,=,9,.,8,d=,9,.,8,所以甲虫的爬行距离,s,与时间,t,的,关系,式,是,s=,9,.,8,t.,(2),当,t=,1,min,=,60,s,时,s=,9,.,8,t=,9,.,8,60,=,588(cm),.,1,2,3,4,5,1,在等差数列,a,n,中,a,1,+a,9,=,10,则,a,5,的值为,(,),.,A.5B.6C.8D.10,解析,:,依题意,得,a,1,+a,9,=,2,a,5,=,10,则,a,5,=,5,故选,A,.,答案,:,A,1,2,3,4,5,2,在等差数列,a,n,中,已知,a,1,=,2,a,2,+a,3,=,13,则,a,4,+a,5,+a,6,等于,(,),.,A.40B.42C.43D.45,解析,:,设等差数列,a,n,的公差为,d,则由,a,2,+a,3,=,13,得,2,a,1,+,3,d=,13,.,a,1,=,2,d=,3,a,4,+a,5,+a,6,=,3,a,5,=,3(,a,1,+,4,d,),=,42,故选,B,.,答案,:,B,1,2,3,4,5,3,已知数列,8,a,2,b,c,是等差数列,则,a,b,c,的值分别是,.,解析,:,依据等差中项的定义,且,8,a,2,是等差数列,得,2,a=,8,+,2,解,得,a=,5,.,由,a,2,b,是等差数列,得,2,2,=a+b.,同理,由,2,b,c,是等差数列,得,2,b=,2,+c.,联,立,解得,a=,5,b=-,1,c=-,4,.,答案,:,5,-,1,-,4,1,2,3,4,5,4,若等差数列,a,n,的前三项依次为,a,2,a+,1,4,a+,2,则它的第五项为,.,解析,:,由题意,知,2,a+,1,是,a,与,4,a+,2,的等差中项,即,2,a+,1,解得,a=,0,故数列,a,n,的前三项依次为,0,1,2,则,a,5,=,0,+,4,1,=,4,.,答案,:,4,1,2,3,4,5,5,夏季高山上的温度从山脚起,每升高,100 m,降低,0,.,7,已知山顶处的温度是,14,.,8,山脚处的温度为,26,此,山相对于山脚处的高度是多少米,?,解,:,每升高,100,m,温度降低,0,.,7,该处温度的变化是等差数列问题,.,设山脚处温度为首项,a,1,=,26,山顶处温度为末项,a,n,=,14,.,8,则,26,+,(,n-,1),(,-,0,.,7),=,14,.,8,解得,n=,17,.,故此山相对于山脚处的高度为,(17,-,1),100,=,1,600(m),.,答,:,此山相对于山脚处的高度为,1,600,m,.,