Bulgarian Mathematical Competitions

Solution:

a) Since $a_{n+1}+b_{n+1}=2 b_{n}-a_{n}+2 a_{n}-b_{n}=a_{n}+b_{n}$, we have
$$a_{n+1}=2\left(a_{n}+b_{n}\right)-3 a_{n}=2\left(a_{1}+b_{1}\right)-3 a_{n}$$

b) Using a), we obtain
$$a_{n+1}-\frac{a_{1}+b_{1}}{2}=-3\left(a_{n}-\frac{a_{1}+b_{1}}{2}\right)$$
whence
$$a_{n+1}-\frac{a_{1}+b_{1}}{2}=(-3)^{n}\left(a_{1}-\frac{a_{1}+b_{1}}{2}\right)$$
Since $\lim _{n \rightarrow \infty} 3^{n}=+\infty$, it follows that if $a_{1}>b_{1}$, then $\lim _{n \rightarrow \infty} a_{2 n}=-\infty$, a contradiction. Analogously, we see that it is not possible to have $a_{1}<b_{1}$. Therefore $a_{1}=b_{1}$.