Question

等差数列{an}的公差为d不等于0,且a1,a3,a9成等比数列,求a1+a3+a9/a2+a4+a10的值. 要详解答案！

Answer 1

设公差为d,
a3=a1+2d,a9=a1+8d
a3²=(a1+2d)²=a1²+4a1d+4d²
a1a9=a1²+8a1d
4a1d+4d²=8a1d
d不为0，所以a1=d
(a1+a3+a9)/(a2+a4+a10)=(a1+3a1+9a1)/(2a1+4a1+10a1)=13/16

Answer 2

设a1=a，按等差数列简化题设：(3a+2d+8d)/(3a+d+3d+9d)=(3a+10d)/(3a+13d) （1）
等比数列a1/a3=a3/a9，即a/(a+2d)=(a+2d)/(a+8d)
=> |a|=|a-2d|
∵d≠0 ∴ a=-(a-2d)
=> a=d
代入（1）答案=13/16

Answer 3

解：qa3=a1+2d
a9=a1+8d有a1 a3 a9比数列qa1=a1+2d
q(a1+2d)=a1+8d联立a1=d
：分子=d+3d+9d=13d分母=2d+4d+10d=
16d 得 13/16