等差数列中a5+a8=18,求a2+a3+a10+a11=?

2024-12-05 05:39:02
推荐回答(3个)
回答1:

a3+a10=a5+a8=18
a2+a11=a5+a8=18
所以原式=18+18=36

回答2:

a2+a3+a10+a11
=(a5-3d)+(a5-2d)+(a8+2d)+(a8+3d)
=2(a5+a8)
=2×18
=36

回答3:

a1+a12=a3+a10=a5+a8
则原式=36