To describe the sequence \( \frac{1}{9}, \frac{1}{6}, \frac{1}{4} \), we first need to determine the pattern between the terms.
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Identify the sequence:
- The first term \( (n=1) \) is \( \frac{1}{9} = 0.1111\ldots \)
- The second term \( (n=2) \) is \( \frac{1}{6} = 0.1666\ldots \)
- The third term \( (n=3) \) is \( \frac{1}{4} = 0.25 \)
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Find a common pattern:
- The denominators of the fractions show a pattern:
- \( 9 \) is \( 3^2 \)
- \( 6 \) is \( 2 \times 3 \)
- \( 4 \) is \( 2^2 \)
However, for a better consistent arithmetic operation, let's express the fractions as decimals for easier examination.
- \( a_1 = 0.1111\ldots \)
- \( a_2 = 0.1666\ldots \)
- \( a_3 = 0.25 \)
- The denominators of the fractions show a pattern:
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Observe the differences:
- The difference from the first to the second term is \( \frac{1}{6} - \frac{1}{9} = \frac{3-2}{18} = \frac{1}{18} \)
- The difference from the second to the third term is \( \frac{1}{4} - \frac{1}{6} = \frac{3-2}{12} = \frac{1}{12} \)
We notice that the denominator increases as \( n \) increases.
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Setting a general formula:
- The denominators seem to relate to square terms or recursive elements.
- The fractions seem to converge towards a simpler formula while they increase steadily.
Let's express them using \( n \):
A potential formula for the \( n \)-th term can be expressed as:
\[ a_n = \frac{1}{9 - 3(n-1)} \]
Therefore:
\[ a_n = \frac{1}{12 - 3n} \]
Hence, using this formula style, we write:
\[ a_n = \frac{1}{12 - 3(n-1)} \]
To conform to the requested format:
\[ a_n = \frac{1}{( 12 - 3(n-1) )} \]
Thus the final answer in the specified format will be:
\[ \boxed{a_n = \frac{1}{(12 - 3(n-1))}} \]