6. The sum of $n$ terms of the geometric progression $\left\{b_{n}\right\}$ is 8 times less than the sum of their reciprocals. Find the product $b_{1} b_{n}$.
Solution: Let $S_{n}=b_{1}+b_{2}+\ldots+b_{n}, S_{n}^{1}=\frac{1}{b_{1}}+\frac{1}{b_{2}}+\ldots+\frac{1}{b_{n}}$. Then $b_{1} b_{n} S_{n}^{1}=\frac{b_{1} b_{n}}{b_{1}}+\frac{b_{1} b_{n}}{b_{2}}+\ldots+\frac{b_{1} b_{n}}{b_{n}}=b_{n}+b_{n-1}+\ldots+b_{1}=S_{n}$. From this we obtain: $b_{1} b_{n}=\frac{S_{n}}{S_{n}^{1}}=\frac{1}{8}$. Answer: $1 / 8$.