等差数列{an}的公差为d,则数列a1+a4,a2+a5,a3+a6,.是什么数列

来源：学生作业帮编辑：作业帮分类：数学作业时间：2024/05/16 08:14:57

考察一般项：
an+a(n+3)=a1+(n-1)d+a1+(n+2)d=2a1+(2n+1)d
a1+a4=2a1+3d
[a(n+1)+a(n+1+3)]-[an+a(n+3)]
=[2a1+[2(n+1)+1]d]-[2a1+(2n+1)d]
=2d,为定值.
数列是以2a1+3d为首项,2d为公差的等差数列.

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