Question

在等差数列{an}中,若a4+a6+a8+a10+a12=120,则a9-1/3a11=?

一定要过程,不要只有答案!!!!!!!

Answer 1

5a1+3d+5d+7d+9d+11d=120
5a1+35d=120
a1+7d=24

a9+1/3a11
=2a1+8d+10d/3
=2a1+14d/3
=2(a1+7d)/3
=2*8
=16

Answer 2

a4+a4+2d+a4+4d+a4+6d+a4+8d=1205a4+20d=120 a4+4d=24 a1+7d=24 a8=24
a9-1/3a11=a8+d-1/3(a8+3d)=2/3a8=16