因为数列{an}是等差数列,所以2a(n+1)=an+a(n+2).
因为数列{1/an}是等差数列,所以2/a(n+1)
=1/an+1/a(n+2)
=[a(n+2)+an]/[an*a(n+2)](通分)
=2a(n+1)/[an*a(n+2)]
所以1/a(n+1)=a(n+1)/[an*a(n+2)],即[a(n+1)]^2=an*a(n+2).
因为数列{an}是等差数列,所以an=a(n+1)-d,a(n+2)=a(n+1)+d.
所以[a(n+1)]^2=an*a(n+2)=[a(n+1)-d]*[a(n+1)+d]=[a(n+1)]^2-d^2.
则d^2=0,d=0,即任意一个非零的常数数列均满足题意.