, 2013

Let $(a, b, c, n)$ be a quadruplet of positive integers with $a<b<c$ such that each of $a, b, c, a+n, b+n, c+2n$ is a term of the Fibonacci sequence, and let
$$b+n=F_{k}$$
for some positive integer $k$. Because $b<b+n$ and $a+n<b+n$, we have $\max \{b, a+n\} \leq F_{k-1}$.
Assume $\min \{b, a+n\} \leq F_{k-2}$. We have $a+n+b \leq F_{k-1}+F_{k-2}=F_{k}=b+n$, which is impossible since $a>0$. Therefore,
$$\begin{gathered} a+n=b=F_{k-1}, \\ n=(b+n)-b=F_{k}-F_{k-1}=F_{k-2}, \end{gathered}$$
and
$$a=(a+n)-n=F_{k-1}-F_{k-2}=F_{k-3} .$$
Let $c+2n=F_{m}$. We have $F_{k} \leq c \leq F_{m-1}$ and therefore $F_{m-2} \leq 2n= 2F_{k-2} \leq F_{k}$.
If $F_{m-2}=F_{k}$ then $F_{k-2}=F_{k-1}=1$ and $a=F_{k-3}=0$ which is impossible. Therefore
$$c=b+n=F_{m-1}=F_{k},$$
and
$$c+2n=F_{m}=F_{k+1}$$
But
$$2n=(c+2n)-c=F_{k+1}-F_{k}=F_{k-1}=a+n .$$
We deduce that
$$F_{k-3}=a=n=F_{k-2}=1 .$$
Hence, $k=4$, $(a, b, c, n)=(1,2,3,1)$ and we check easily that
$$a=F_{1},\ b=a+n=F_{3},\ c=b+n=F_{4}\ \text{ and }\ c+2n=F_{5} .$$