已知数列{an}是等差数列,其前n项和为Sn,a3=7,S4=24

分类: 作业答案 • 2021-12-31 20:26:16

问题描述：

已知数列{an}是等差数列,其前n项和为Sn,a3=7,S4=24
求数列{an}的通项公式

答

a3 = a1 + 2d = 7
S4 = (a4 + a1)*4/2 = (a3 +d +a1)*2 = (7 +a1 +d)*2 = 24
a1 + d = 5
d = 2
a1 = 3
an = 3 + 2(n-1) = 2n+1