Question

设等差数列{An}的公差为－2，且A1+A4+A7+...+A97=50. 求A3+A6+A9+.

设等差数列{An}的公差为－2，且A1+A4+A7+...+A97=50. 求A3+A6+A9+...+A99=？

Answer 1

A3+A6+A9+...+A99
=(A1+2d)+(A4+2d)+(A7+2d)+...+(A97+2d)
=(A1+A4+A7+...+A97)+(2d+2d+2d+...+2d) (一共33个)
=50+(-2*2)*33
=50-132
=-82

追问

谢谢

Answer 2

an = a1+(n-1)d, d=-2

a1+a4+a7+...+a97 =50
(a1+a97)33/2 =50
(2a1+96d)33=100
a1= (100/33 + 192)/2
= 3218/33

a3+a6+....+a99
= (a3+a99)33/2
=( 2a1+ 100d)33/2
=-82