Ukrajina 2008

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.