Question

已知数列an,bn都是等差数列，且满足a1+a2+a3...+an\b1+b2+....+bn=7n+2/n+3求a5/b5

Answer 1

令n=5

则(a1+……++a9)/(b1+……+b9)=65/12
[(a1+a9)*9/2]/[(b1+b9)*9/2]=65/12
(a1+a9)/(n1+n9)=65/12
2a5/2b5=65/12
a5/b5=65/12

Answer 2

解：
(a1+a2+...+a9)/(b1+b2+...+b9)
=[(a1+a9)+(a2+a8)+(a3+a7)+(a4+a6)+a5]/[(b1+b9)+(b2+b8)+(b3+b7)+(b4+b6)+b5]
=[(2a5)×4+a5]/[(2b5)×4+b5]
=(9a5)/(9b5)
=a5/b5
=(7×9+2)/(9+3)
=65/12

a5/b5=65/12