China Mathematical Competition

Let the common difference of $\{a_n\}$ be $d$ and the common ratio of $\{b_n\}$ be $q$. Then $$3+d=q,\stackrel{◯}{1}$$ $$3(3+4d)=q^2.\stackrel{◯}{2}$$ Substituting $\stackrel{◯}{1}$ into $\stackrel{◯}{2}$, we have $9+12d=d^2+6d+9$. Then we get $d=6$ and $q=9$. Therefore, $3+6(n-1)={\log }_a{9}^{n-1}+\beta$ or $6n-3=(n-1){\log }_{a}9+\beta$ holds for every positive integer $n$. Letting $n=1$ and $n=2$ in turn, we find that $\alpha =\sqrt[3]{3}$ and $\beta =3$. Consequently, $\alpha +\beta =\sqrt[3]{3}+3$. $\square$