已知{an}为公差为—2的等差数列,若a3+a6+a9+..+a99=—82,则a1+a4+a7+..a97=?

题目

已知{an}为公差为—2的等差数列,若a3+a6+a9+..+a99=—82,则a1+a4+a7+..a97=?

答案

a3+a6+a9+…+a99
=(a1+2d)+(a4+2d)+(a7+2d)+…+(a97+2d)
= a1+a4+a7+…+a97+(99/3)*2d
= a1+a4+a7+…+a97+66d
=-82
a1+a4+a7+…+a97=-82-66d=-82-66*(-2)=50