Question

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

Answer 1

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.