Question: e) Sā of 9/5 + 1/10 + 1/20 + 1/40 + ...\n\nThe sum of the first two terms of a geometric sequence is 15, and the sum of the second and third terms is 60. Determine the first three terms of the sequence algebraically.
1. **State the problem:**\nFind the sum of the first 6 terms ($S_6$) of the geometric sequence starting with $\frac{9}{5}, \frac{1}{10}, \frac{1}{20}, \frac{1}{40}, \ldots$.\n\n2. **Identify the sequence:**\nThe given terms are $\frac{9}{5}, \frac{1}{10}, \frac{1}{20}, \frac{1}{40}, \ldots$. We notice each term after the first is half the previous term, so the common ratio $r = \frac{1}{2}$.\n\n3. **Formula for sum of first $n$ terms of geometric sequence:**\n$$S_n = a \frac{1-r^n}{1-r}$$\nwhere $a$ is the first term and $r$ is the common ratio.\n\n4. **Calculate $S_6$:**\nGiven $a = \frac{9}{5}$ and $r = \frac{1}{2}$,\n$$S_6 = \frac{9}{5} \times \frac{1 - (\frac{1}{2})^6}{1 - \frac{1}{2}}$$\n\n5. **Simplify denominator:**\n$$1 - \frac{1}{2} = \frac{1}{2}$$\n\n6. **Calculate numerator:**\n$$(\frac{1}{2})^6 = \frac{1}{64}$$\nSo,\n$$1 - \frac{1}{64} = \frac{63}{64}$$\n\n7. **Substitute back:**\n$$S_6 = \frac{9}{5} \times \frac{\frac{63}{64}}{\frac{1}{2}} = \frac{9}{5} \times \frac{63}{64} \times 2$$\n\n8. **Multiply:**\n$$S_6 = \frac{9}{5} \times \frac{126}{64} = \frac{9}{5} \times \frac{63}{32}$$\n\n9. **Final multiplication:**\n$$S_6 = \frac{9 \times 63}{5 \times 32} = \frac{567}{160}$$\n\n10. **Simplify fraction if possible:**\n$567$ and $160$ share no common factors other than 1, so\n$$S_6 = \frac{567}{160}$$\n\n**Answer:** The sum of the first 6 terms is $$\boxed{\frac{567}{160}}$$.