等差数列的应用(写过程)(1) a1=105,an=994,d=7,求Sn(2) an=8n+2,d=5,求S20(3) d=1/3,n=37,Sn=629,求a1及an
来源:学生作业帮助网 编辑:作业帮 时间:2024/06/24 11:06:51
![等差数列的应用(写过程)(1) a1=105,an=994,d=7,求Sn(2) an=8n+2,d=5,求S20(3) d=1/3,n=37,Sn=629,求a1及an](/uploads/image/z/9493691-59-1.jpg?t=%E7%AD%89%E5%B7%AE%E6%95%B0%E5%88%97%E7%9A%84%E5%BA%94%E7%94%A8%EF%BC%88%E5%86%99%E8%BF%87%E7%A8%8B%EF%BC%89%EF%BC%881%EF%BC%89+a1%3D105%2Can%3D994%2Cd%3D7%2C%E6%B1%82Sn%EF%BC%882%EF%BC%89+an%3D8n%2B2%2Cd%3D5%2C%E6%B1%82S20%EF%BC%883%EF%BC%89+d%3D1%2F3%2Cn%3D37%2CSn%3D629%2C%E6%B1%82a1%E5%8F%8Aan)
等差数列的应用(写过程)(1) a1=105,an=994,d=7,求Sn(2) an=8n+2,d=5,求S20(3) d=1/3,n=37,Sn=629,求a1及an
等差数列的应用(写过程)
(1) a1=105,an=994,d=7,求Sn
(2) an=8n+2,d=5,求S20
(3) d=1/3,n=37,Sn=629,求a1及an
等差数列的应用(写过程)(1) a1=105,an=994,d=7,求Sn(2) an=8n+2,d=5,求S20(3) d=1/3,n=37,Sn=629,求a1及an
(1)an=a1+(n-1)xd,代入994=105+(n-1)x7,求得n=128,所以sn=(a1+an)xn/2
=(105+994)x128/2=70336
(2)a1=10,a20=162,s20=(10+162)x20/2=1720
(3)37a1+37x36x1/3x1/2=629,求得a1=11,a37=11+36x1/3=23