是否存在数列{an}同时满足下列条件
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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}

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