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Algebra Difficulty 4.8 AIME Prove it Ukraine

Sequence $\{f_n\}$ is defined as follows: $f_1 = 1$ , $f_2 = 2$ , $f_{n+2} = f_{n+1} + f_n$ for random natural $n$ . What may be the greatest number of members of the sequence $\{f_n\}$ among the consecutive members of the increasing arithmetic progression?

Solution

Let the first consecutive members be $f_1 = a$ , $f_m = a + d$ . Since $a > 0$ , $d > 0$ , $f_{m+1} > f_m$ , $f_{m+2} = f_{m+1} + f_m > 2a + 2d > a + 2d$ , the third member may only be $f_{m+1}$ .

So $f_{m+1} = a + 2d$ . It's clear that the next member of progression does not satisfy the condition because $f_{m+2} = 2a + 3d > a + 3d$ . But we can easily find the three consecutive members 1, 2, 3, and 2, 5, 8.

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