题目

在等差数列｛an}中:(1)若S12=21求a2+a8+a11（2）若a7/a4=2求a13/a7（3）若a5=9,S6=36,求a7+a8+a9

提问时间：2021-12-26

答案

S12=21求a2+a8+a11
s12=6(a6+a7)
21=7(a6+a7)
a6+a7=3
a2+a5+a8+a11
=2(a6+a7)
=2*3
=6
a7/a4=2
(a4+3d)/a4=2
1+3d/a4=2
3d/a4=1
a4=3d
a13/a7
=(a4+9d)/(a4+3d)
=(3d+9d)/(3d+3d)
=12/6
=2
s6=(a2+a5)*3
36=3(a2+9)
12=a2+9
a2=3
a5=a2+3d
9=3+3d
d=2
a7+a8+a9
=3a2+(5d+6d+7d)
=3a2+18d
=3*3+18*2
=9+36
=45