To determine which of the given sequences is a geometric sequence, we need to check if the ratio between consecutive terms is constant for each sequence.
### Sequence A: [tex]\(1, 3, 9, 27, \ldots\)[/tex]
1. To find the common ratio, we divide each term by the preceding term:
- [tex]\(\frac{3}{1} = 3\)[/tex]
- [tex]\(\frac{9}{3} = 3\)[/tex]
- [tex]\(\frac{27}{9} = 3\)[/tex]
Since the ratio between consecutive terms is constant and equals 3, sequence A is a geometric sequence.
### Sequence B: [tex]\(3, 6, 9, 12, \ldots\)[/tex]
1. To find the common ratio, we divide each term by the preceding term:
- [tex]\(\frac{6}{3} = 2\)[/tex]
- [tex]\(\frac{9}{6} = 1.5\)[/tex]
- [tex]\(\frac{12}{9} = \frac{4}{3} \approx 1.33\)[/tex]
Since the ratios are not constant, sequence B is not a geometric sequence.
### Sequence C: [tex]\(1, \frac{1}{2}, \frac{1}{6}, \frac{1}{24}, \ldots\)[/tex]
1. To find the common ratio, we divide each term by the preceding term:
- [tex]\(\frac{\frac{1}{2}}{1} = \frac{1}{2}\)[/tex]
- [tex]\(\frac{\frac{1}{6}}{\frac{1}{2}} = \frac{1/6}{1/2} = \frac{1}{6} \times \frac{2}{1} = \frac{2}{6} = \frac{1}{3}\)[/tex]
- [tex]\(\frac{\frac{1}{24}}{\frac{1}{6}} = \frac{1/24}{1/6} = \frac{1}{24} \times \frac{6}{1} = \frac{6}{24} = \frac{1}{4}\)[/tex]
Since the ratios between terms are not constant, sequence C is not a geometric sequence.
### Sequence D: [tex]\(-1, -1, 1, -1, -1, 1, \ldots\)[/tex]
1. To find the common ratio, we divide each term by the preceding term:
- [tex]\(\frac{-1}{-1} = 1\)[/tex]
- [tex]\(\frac{1}{-1} = -1\)[/tex]
- [tex]\(\frac{-1}{1} = -1\)[/tex]
- [tex]\(\frac{-1}{-1} = 1\)[/tex]
- [tex]\(\frac{1}{-1} = -1\)[/tex]
Since the ratios are not constant (alternating between 1 and -1), sequence D is not a geometric sequence.
### Conclusion
- Sequence A ([tex]\(1, 3, 9, 27, \ldots\)[/tex]) is a geometric sequence.
- Sequences B, C, and D are not geometric sequences.
Thus, sequence A is the geometric sequence.
