Austria 2010

Since $a_k=a+kd$, we have $b_n=(n+1)\cdot a+\frac{n(n+1)}{2}\cdot d$. If we assume $b_n=2010$, we therefore obtain $$a=\frac{4020-n(n+1)d}{2(n+1)}=\frac{2010}{n+1}-\frac{dn}{2}.$$ Since both $a$ and $d$ are positive, we see from the first fraction, that $n(n+1)<4020$ must hold. We therefore have $n(n+1)<4020<4225=65^2$, and thus $n<65$. Furthermore, $n+1$ must divide $4020$. Since $4020=3\cdot 4\cdot 5\cdot 67$ holds, the largest divisor of $4020$ less than $65$ is $60$. The largest possible value for $n+1$ is therefore $60$, and we have $n=59$.

Substituting this value yields $a=\frac{67-59d}{2}$. We see that $d$ must be odd and less than $2$, and we therefore have the unique solution $d=1$, which yields $a=4$.

The only possible sequence $⟨a_n⟩$, for which $b_{59}=2010$ is therefore the sequence $⟨4,5,6,\dots ⟩$.

qed