Question

数列an是等差数列，a1,a2,a3成等比数列。则a1+a3+a9/a2+a4+a10等于多少？

Answer 1

解；设首相为a
则a1=a,a3=a+2d,a9=a+8d,a2=a+d,a4=a+3d,a10=a+9d
a1，a3，a9成等比数列
(a+2d)^2=a*(a+8d)
a=d
所以（a1+a3+a9）/（a2+a4+a10）＝(a+a+2d+a+8d)/(a+d+a+3d+a+9d)
=13a/16a=13/16

Answer 2

等差数列中，a1=a2-d，a3=a2+d。a1、a2、a3等比，则(a2)^2=a1xa3=(a2-d)(a2+d)=(a2)^2-d^2，d^2=0，d=0！与公差不为零的已知条件矛盾吧？

Answer 3

a2=a1+d,a3=a1+2d,由a1,a2,a3成等比数列，所以可推出a1*（a1+2d）=（a1+d）^2,所以d=0，所以a1=a3=a9=a2=a4=a10，所以答案为1/3

Answer 4

设等差d
a2=a1+d, a3=a1+2d
(a1+d)^2=a1(a1+2d)
2a1d+d^2=2a1d
d=0
an=a1
（a1+a3+a9）/（a2+a4+a10）=3a1/(3a1) =1