Estonian Mathematical Olympiad

a. Let the common ratio of the progression be $q$. W.l.o.g., assume that $q>1$. One can also assume that the first term of the progression is $3$. Then $45=3\cdot q^k$ and $2025=3\cdot q^l$, where $k$ and $l$ are integers. This implies $q^k=15$ and $q^l=675$. Therefore $q^{kl}=15^l$, as well as $q^{kl}=675^k$. Since $15=3\cdot 5$ and $675=3^3\cdot 5^2$, the equality $15^l=675^k$ implies $3^l\cdot 5^l=3^{3k}\cdot 5^{2k}$ which simplifies to $5^{l-2k}=3^{3k-1}$. As $k$ and $l$ are integers, this is possible only if $l-2k=3k-l=0$. But then $k=(l-2k)+(3k-l)=0$ which is obviously false.

b. If $q=\frac{15}{\sqrt{3}}$ then the next term after $3$ is $\frac{45}{\sqrt{3}}$ and the term after the next term is $\frac{3\cdot 15^3}{5}=3^4\cdot 5^2=2025$. Hence there exists a suitable geometric progression.