设数列{an}为等差数列,首项为a1,公差为d,则有an-a(n-1)=d,d≠0
1/an-1/a(n-1)
=[a(n-1)-an]/[ana(n-1)]
=-d/[ana(n-1)]
若要{1/an}为等差数列,则-d/[ana(n-1)]为定值.
ana(n-1)
=[a1+(n-1)d][a1+(n-2)d]
=a1^2+(n-2)a1d+(n-1)a1d+(n-1)(n-2)d^2
=d^2n^2+(2a1d-3d^2)n-3a1d+2d^2+a1^2
要此多项式为定值,即与n无关,则需
d^2=0 2a1d-3d^2=0
又d≠0,因此找不到满足题意的d
不存在同时满足上述两条件的数列{an}