Question

已知数列1/an是等差数列,且a1a3+a3a5+a5a1=3/5,a1a3a5=1/15

Answer 1

数列{1/an}是等差数列，设公差为d
2/a3=1/a1+1/a5=(a1+a5)/(a1a5)
a3(a1+a5)=2a1a5
a1a3+a3a5+a5a1
=a3(a1+a5)+a1a5
=2a1a5+a1a5
=3a1a5=3/5
a1a5=1/5
a1a3a5=1/15
a3=(1/15)/(a1a5)=(1/15)/(1/5)=1/3
a3(a1+a5)=2a1a5=2/5
a1+a5=(2/5)/a3=(2/5)/(1/3)=6/5
a1、a5是方程x²-(6/5)x+(1/5)=0的两根。
(x-1)(x-1/5)=0
x=1或x=1/5
(1)
a1=1 a5=1/5时，
1/a5-1/a1=4d=1/(1/5)-1/1=4
d=1 1/an=1+1×(n-1)=n
an=1/n
数列{an}的通项公式为an=1/n
(2)
a1=1/5 a5=1时，
1/a5-1/a1=4d=1/1-1/(1/5)=-4
d=-1 1/an=1/a1+(n-1)d=1/(1/5)+(-1)(n-1)=-(n-6)
an=-1/(n-6)
数列{an}的通项公式为an=-1/(n-6)

Answer 2

(a1a3+a3a5+a5a1)/a1a3a5=(3/5)/(1/15)=9

1/a1+1/a3+1/a5=9
3*(1/a3)=9
a3=1/3