1. **Problem Statement:** Given a geometric series where the 4th term is 10 and the 7th term is 80, find: i. The common ratio $r$ ii. The first term $a$ iii. The sum of the first 20 terms $S_{20}$ 2. **Formula and Rules:** - The $n$th term of a geometric series is given by: $$a_n = a r^{n-1}$$ - The sum of the first $n$ terms is: $$S_n = a \frac{1-r^n}{1-r}$$ for $r \neq 1$ 3. **Find the common ratio $r$:** From the problem: $$a_4 = a r^{3} = 10$$ $$a_7 = a r^{6} = 80$$ Divide the second equation by the first: $$\frac{a r^{6}}{a r^{3}} = \frac{80}{10} \Rightarrow r^{3} = 8$$ Taking cube root: $$r = \sqrt[3]{8} = 2$$ 4. **Find the first term $a$:** Substitute $r=2$ into $a_4 = a r^{3} = 10$: $$a \times 2^{3} = 10 \Rightarrow a \times 8 = 10 \Rightarrow a = \frac{10}{8} = 1.25$$ 5. **Find the sum of the first 20 terms $S_{20}$:** Using the sum formula: $$S_{20} = 1.25 \times \frac{1 - 2^{20}}{1 - 2} = 1.25 \times \frac{1 - 1,048,576}{-1} = 1.25 \times (1,048,575) = 1,310,718.75$$ Rounded to the nearest whole number: $$S_{20} \approx 1,310,719$$ **Final answers:** - Common ratio $r = 2$ - First term $a = 1.25$ - Sum of first 20 terms $S_{20} = 1,310,719$