Question

在等差数列{an}中,若a2+a4+a8+a10=80,则a7-0.5a8等于

Answer 1

设a2=m,a4=m+2d,a8=m+6d,a10=m+8d
m+m+2d+m+6d+m+8d=80
解得m+4d=20
a7-0.5a8=m+5d-0.5（m+6d）=0.5m+2d=10

Answer 2

a2+a4+a8+a10=4a1+20d=4(a1+5d)=4a6=80
所以a6=20
a7-0.5a8=a1+6d-0.5(a1+7d)=0.5a1+2.5d=0.5(a1+5d)=0.5a6=10

Answer 3

=10.
因为a2+a4+a8+a10=80，所以由等差数列性质可知，4a6=80,所以a6=20.
a7-0.5a8=a6+d-0.5(a6+2d)
=a6+d-0.5a6-d
=0.5a6
=10.

Answer 4

由于是等差数列{an}，那么an=a1+（n-1）d
则a2+a4+a8+a10=80可表示为
4a1+20d=80
a1+5d=20
设a7-0.5a8=y
2a7-a8=2y=2a1+12d-7d-a1=a1+5d=20
y=10