Question

公差不为零的等差数列{an}的第二三及第六项构成等比数列,则a1+a3+a5/a2+a4+a6

Answer 1

设首项为a1,公差为d 则 a2=a1+d,a3=a1+2d,a6=a1+5d
由第二三及第六项构成等比数列，得
(a3)^2=a2*a6
(a1+2d)^2=(a1+d)(a1+5d)
解得 d=0(舍掉)，d=-2a1
(a1+a3+a5)/(a2+a4+a6)
=(a1+a1+2d+a1+4d)/(a1+d+a1+3d+a1+5d)
=(3a1+6d)/(3a1+9d)
=(3a1-12a1)/(3a1-18a1)
=3/5

Answer 2

a2=a4-2d,a3=a4-d,a6=a4+2d
第二三及第六项构成等比数列,则
(a3)²=a2*a6 ， 即
(a4-d)(a4-d)=(a4-2d)(a4+2d)
得 a4=2.5d
故 a3=a4-d=1.5d ；

因 a1+a5=2a3，a2+a6=2a4
故
(a1+a3+a5)/(a2+a4+a6)
=a3/a4
=1.5d/2.5d
=0.6