1. **Problem statement:** We are given a geometric progression (GP) with terms $b_2 = \frac{1}{2}$ and $b_6 = 128$, and the common ratio $q > 0$. We need to find the sum of the first 7 terms of this GP. 2. **Recall the formula for the $n$-th term of a GP:** $$b_n = b_1 q^{n-1}$$ where $b_1$ is the first term and $q$ is the common ratio. 3. **Use the given terms to form equations:** $$b_2 = b_1 q = \frac{1}{2}$$ $$b_6 = b_1 q^5 = 128$$ 4. **Divide the second equation by the first to eliminate $b_1$:** $$\frac{b_6}{b_2} = \frac{b_1 q^5}{b_1 q} = q^4 = \frac{128}{\frac{1}{2}} = 128 \times 2 = 256$$ 5. **Solve for $q$:** $$q^4 = 256$$ Since $q > 0$, $$q = \sqrt[4]{256} = 4$$ 6. **Find $b_1$ using $b_2 = b_1 q$:** $$b_1 = \frac{b_2}{q} = \frac{\frac{1}{2}}{4} = \frac{1}{8}$$ 7. **Sum of the first 7 terms of a GP:** $$S_7 = b_1 \frac{q^7 - 1}{q - 1}$$ 8. **Calculate $S_7$:** $$S_7 = \frac{1}{8} \times \frac{4^7 - 1}{4 - 1} = \frac{1}{8} \times \frac{16384 - 1}{3} = \frac{1}{8} \times \frac{16383}{3} = \frac{16383}{24}$$ 9. **Final answer:** $$S_7 = \frac{16383}{24}$$