Question

已知{an}是等差数列,且公差d≠0,又a1,a2,a4依次成等比数列,则a1+a4+a10/a2+a4+a7=

求解答过程，写的详细些，谢谢了。

Answer 1

∵{an}是等差数列
∴a2=a1+d
a4=a1+3d
∵a1,a2,a4依次成等比数列
∴(a2)^2=a1*a4
即（a1+d)^2=a1(a1+3d)
a1^2+2a1+d^2=a1^2+3a1d
a1d=d^2
∵公差d≠0,
∴a1=d
∴a4=4a1 a7=7a1 a10=10a1 a2=2a1
式子要是有括号是（a1+4a1+10a1）/（2a1+4a1+7a1）=15a1/13a1=15/13
式子要是没有括号是a1+4a1+10a1/2a1+4a1+7a1=5a1+5+11a1=16a1+5

Answer 2

15/13
a4=a1+3d，a2=a1+d带入a1*a4=a2*a2（a1,a2,a4 依次成等比数列）
整理得a1=d
a1+a4+a10=3a1+12d=15d
a2+a4+a7=3a1+10d=13d
所以（a1+a4+a10）/（a2+a4+a7）=15/13

Answer 3

因为a1,a2,a4成等比，a2*a2=a1*a4即（a1+d）^2=a1*(a1+3d),
又d≠0，故a1=d;
a1+a4+a10/a2+a4+a7=15d/13d=15/13;

Answer 4

因为a1,a2,a4成等比数列，所以a2^2=a1a4,又因为a2=a1+d,a4=a1+3d,因为d不等于0，所以d=a1,原式=（15a1)/(13a1)=15/13.