jmc

We claim that if $F_a,$ $F_b,$ $F_c$ form an increasing arithmetic sequence, then $(a,b,c)$ must be of the form $(n,n+2,n+3)$ for some positive integer $n.$ (The only exception is $(2,3,4).$)

From $F_c-F_b=F_b-F_a,$ we get $$F_c=F_b+(F_b-F_a)<F_b+F_{b+1}=F_{b+2}.$$Also, $F_c>F_b.$ Therefore, $F_c=F_{b+1}.$

Then $$\begin{array}{rl}{F}_a& =2F_b-F_c\\ & =2F_b-F_{b+1}\\ & =F_b-(F_{b+1}-F_b)\\ & =F_b-F_{b-1}\\ & =F_{b-2}.\end{array}$$Then $a$ must be equal to $b-2$ (unless $b=3,$ which leads to the exceptional case of $(2,3,4)$). Taking $n=b-2,$ we get $(a,b,c)=(n,n+2,n+3).$

Then $a+(a+2)+(a+3)=2000,$ so $a=\boxed{665}.$