4. The sum $b_{7}+b_{6}+\ldots+b_{2019}$ of the terms of the geometric progression $\left\{b_{n}\right\}, b_{n}>0$ is 27, and the sum of their reciprocals $\frac{1}{b_{7}}+\frac{1}{b_{6}}+\ldots+\frac{1}{b_{2019}}$ is 3. Find the product $b_{7} \cdot b_{6} \cdot \ldots \cdot b_{2019}$.
Answer: $b_{7} \cdot b_{6} \cdot \ldots \cdot b_{2019}=3^{2013}$.