Question

等差数列{an}a4+a6+a8+a10+a12=120.则a9-1/3a11的值为

Answer 1

a4+a6+a8+a10+a12=120
5a8=120
a8=24
a9-1/3*a11 =a8+d-1/3*(a8+3d)
到这步都没有问题 之后 问题出现了
应该是a8+d-1/3a8-d 因为括号打开了 =a8-1/3a8 =2/3a8=16

Answer 2

a4+a6+a8+a10+a12=120
5a8=120
a8=24

a9-1/3*a11
=a8+d-1/3*(a8+3d)
=a8+d-1/3a8+d
=a8+1/3a8
=24+1/3*24
=24+8
=32