To find the common ratio of the sequence \( \frac{2}{3}, \frac{1}{6}, \frac{1}{24}, \frac{1}{96} \), we can divide any term by the previous term.
Let's calculate the common ratio \( r \):
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From the first term \( \frac{2}{3} \) to the second term \( \frac{1}{6} \): \[ r = \frac{\frac{1}{6}}{\frac{2}{3}} = \frac{1}{6} \times \frac{3}{2} = \frac{3}{12} = \frac{1}{4} \]
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Now, let's check from the second term \( \frac{1}{6} \) to the third term \( \frac{1}{24} \): \[ r = \frac{\frac{1}{24}}{\frac{1}{6}} = \frac{1}{24} \times 6 = \frac{6}{24} = \frac{1}{4} \]
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Lastly, let's check from the third term \( \frac{1}{24} \) to the fourth term \( \frac{1}{96} \): \[ r = \frac{\frac{1}{96}}{\frac{1}{24}} = \frac{1}{96} \times 24 = \frac{24}{96} = \frac{1}{4} \]
Since in each case we found that the common ratio \( r = \frac{1}{4} \), the answer is:
B) \( \frac{1}{4} \)