在等差数列｛an}中,公差d不等于.0,且a1,a2,a5成等比数列,求a1+a3+a9/a2+a4+a10的值
问题描述：
在等差数列｛an}中,公差d不等于.0,且a1,a2,a5成等比数列,求a1+a3+a9/a2+a4+a10的值
答
由a1,a2,a5成等比数列得a2^2=a1*a5,即(a1+d)^2=a1(a1+4d),化简得d=2a1.
a1+a3+a9/a2+a4+a10=[a1+(a1+2d)+(a1+8d)]/[(a1+d)+(a1+3d)+(a1+9d)]=(3a1+10d)/(3a1+13d)=(3a1+20d)/(3a1+26a1)=23/29
答
a1a5=a2^2
a1*(a1+4d)=(a1+d)^2 公差d不等于.0 求解d与a1关系
a1+a3+a9/a2+a4+a10=(3a1+8d)/(3a1+13d)将d与a1关系代入求解化简即可
答
a1 a1+d a1+4d 成等比数列
(a1+d)^2=a1(a1+4d)
a1^2+2a1d+d^2=a1^2+4a1d
d^2=2a1d d不等于0
d=2a1
a1+a3+a9/a2+a4+a10
=(a1+a1+2d+a1+8d)/(a1+d+a1+3d+a1+9d)
=23a1/29a1
=23/29