Question

设 {an}是公差为-2的等差数列，如果a1+a4+a7+......a97=50,那么a3+a6+a9+......a99=?

请写出步骤，谢谢啊。

Answer 1

a1+a4+a7+...+a97=50
一共有（97-1）/3 + 1=33项
a3=a1+2d
a6=a4+2d
……
a99=a97+2d
a3+a6+a9+...+a99=a1+a4+a7+...+a97+2d*33=50+66d=50-66*2=-82

Answer 2

a3+a6+a9+......a99
=(a1-2*2)+9a4-2*2)+(a7-2*2)+......(a97-2*2)
=50-2*2*33=-82

Answer 3

a1比a3大4
所以=50-4*33=-82