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,等差数列通项公式,问题 某工厂的仓库里堆放一批钢管,共堆放了,7,层,,试从上到下列出每层钢管的数量,.,引入,每层钢管数为,4,,,5,,,6,,,7,,,8,,,9,,,10,新授,等差数列,一般地,如果一个数列从第二项起,每一项与它前,一项的差等于同一个常数,这个数列就叫做等差数列,这个常数就叫做等差数列的公差(常用字母,“,d,”,表示),练习一,抢答:下列数列是否为等差数列?,1,,,2,,,4,,,6,,,8,,,10,,,12,,,0,,,1,,,2,,,3,,,4,,,5,,,6,,,3,,,3,,,3,,,3,,,3,,,3,,,3,,,2,,,4,,,7,,,11,,,16,,,8,,,6,,,4,,,-2,,,0,,,2,,,4,,,3,,,0,,,3,,,6,,,9,,,练习二,说出下列等差数列的公差,0,,,1,,,2,,,3,,,4,,,5,,,6,,,3,,,3,,,3,,,3,,,3,,,3,,,3,,,-8,,,-6,,,-4,,,-2,,,0,,,2,,,4,,,3,,,0,,,-3,,,-6,,,-9,,,d,=1,d,=0,d,=2,d,=,-,3,常数列,新授,根据等差数列的定义填空,a,2,a,1,d,,,a,3,d,(),d,a,1,d,,,a,4,d,(),d,a,1,d,,,a,n,d,a,2,a,1,+,d,2,a,3,a,1,+2,d,3,a,1,(,n,1,),等差数列的通项公式,例,1,求等差数列,8,,,5,,,2,,,的通项公式和第,20,项,解 因为,a,1,8,,,d,5,8,3,,,所以这个数列的通项公式是,a,n,=8,(,n,1)(,3),,,即,a,n,3,n,11,所以,a,20,320,11,49.,新授,例,2,等差数列,5,,,9,,,13,,,的第多少项是,401,?,解 因为,a,1,5,,,d,9,(,5),4,,,a,n,401,,,所以,401,5,(,n,1)(,4),解得,n,100,即这个数列的第,100,项是,401,新授,练习三,(,1,)求等差数列,3,,,7,,,11,,,的第,4,,,7,,,10,项;,(,2,)求等差数列,10,,,8,,,6,,,的第,20,项,练习四,在等差数列,a,n,中:,(,1,),d,,,a,7,8,,求,a,1,;,(,2,),a,1,12,,,a,6,27,,求,d,例,3,在,3,与,7,之间插入一个数,A,,使,3,,,A,,,7,成等差数列,解 因为,3,,,A,,,7,成等差数列,,所以,A,3,7,A,,,2,A,3,7,解得,A,5,一般地,如果,a,,,A,,,b,成等差数列,那么,A,叫做,a,与,b,的等差中项,A,a,+,b,2,新授,练习五,求下列各组数的等差中项:,(,1,),732,与,136,;,(,2,)与,42,新授,例,4,已知一个等差数列的第,3,项是,5,,第,8,项是,20,,,求它的第,25,项,解 因为,a,3,5,,,a,8,20,,,根据通项公式得,整理,得,解此方程组,得,a,1,1,,,d,3,所以,a,25,1,(25,1)3,71.,a,1,(3,1),d,5,a,1,(8,1),d,20,a,1,2,d,5,a,1,7,d,20,练习六,(,1,),已知等差数列,a,n,中,,a,1,=3,,,a,n,=21,,,d,=2,,,求,n,(,2,),已知等差数列,a,n,中,,a,4,=10,,,a,5,=6,,,求,a,8,和,d,新授,例,5,梯子的最高一级是,33 cm,,最低一级是,89 cm,,,中间还有,7,级,各级的宽度成等差数列,求中间各级的宽度,解 用,a,n,表示题中的等差数列,已知,a,1,=33,,,a,n,=89,,,n,=9,,,则,a,9,=33+,(,9,1,),d,,即,89=33+8,d,,,解得,d,=7,于是,a,2,=33+7=40,,,a,3,=40+7=47,,,a,4,=47+7=54,,,a,5,=54+7=61,,,a,6,=61+7=68,,,a,7,=68+7=75,,,a,8,=75+7=82,即梯子中间各级的宽从上到下依次是,40 cm,,,47 cm,,,54 cm,,,61 cm,,,68 cm,,,75 cm,,,82 cm,新授,例,6,已知一个直角三角形的三条边的长度成等差数列,求证:它们的比是,345,证明设这个直角三角形的三边长分别为,a,d,,,a,,,a,d,根据勾股定理,得,(,a,d,),2,a,2,(,a,d,),2,解得,a,=4,d,于是这个直角三角形的三边长是,3,d,,,4,d,,,5,d,,,即这个直角三角形的三边长的比是,345,1,等差数列的定义及通项公式,2.,等差中项的定义及公式,3,等差数列定义、通项公式和中项公式的应用,归纳小结,
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