Question

Write an equation to describe the sequence below.Use n to represent the position of a term in the sequence,where n = 1 for the first term.
1/9 , 1/6 , 1/4 ,
Write your answer using decimals and integers.
an = __ (__)n-1

Answer 1

To describe the sequence \( \frac{1}{9}, \frac{1}{6}, \frac{1}{4} \), we first need to determine the pattern between the terms.

1. **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 \( \frac{1}{4} = 0.25 \)

2. **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.
- \( a_1 = 0.1111\ldots \)
- \( a_2 = 0.1666\ldots \)
- \( a_3 = 0.25 \)

3. **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.

4. **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:

\[
\boxed{a_n = \frac{1}{(12 - 3(n-1))}}
\]