Step 1: Determine the Type of Sequence

The given terms are:

$a_1 = \frac{1}{3}, \quad a_2 = \frac{1}{6}, \quad a_3 = \frac{1}{12}$

Test for Arithmetic Sequence:

In an arithmetic sequence, the difference between consecutive terms is constant: $d = a_2 - a_1$

Compute $a_2 - a_1$: $d = \frac{1}{6} - \frac{1}{3} = \frac{1}{6} - \frac{2}{6} = -\frac{1}{6}$

Now compute $a_3 - a_2$: $d = \frac{1}{12} - \frac{1}{6} = \frac{1}{12} - \frac{2}{12} = -\frac{1}{12}$

Since the differences are not equal ($-\frac{1}{6} \neq -\frac{1}{12}$), the sequence is not arithmetic.

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Test for Geometric Sequence:

In a geometric sequence, the ratio between consecutive terms is constant: $r = \frac{a_2}{a_1}$

Compute $\frac{a_2}{a_1}$: $r = \frac{\frac{1}{6}}{\frac{1}{3}} = \frac{1}{6} \cdot \frac{3}{1} = \frac{1}{2}$

Now compute $\frac{a_3}{a_2}$: $r = \frac{\frac{1}{12}}{\frac{1}{6}} = \frac{1}{12} \cdot \frac{6}{1} = \frac{1}{2}$

Since the ratios are equal ($\frac{1}{2}$), the sequence is geometric.

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Step 2: Find $a_1$ and $r$

• The first term $a_1 = \frac{1}{3}$.
• The common ratio $r = \frac{1}{2}$.

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Final Answer:

1. The sequence is geometric.
2. $a_1 = \frac{1}{3}$
3. $r = \frac{1}{2}$

Would you like a detailed explanation of how geometric sequences work or have any other questions?

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Related Questions:

1. What is the formula for the $n$-th term of a geometric sequence?
2. How can you verify whether a sequence is arithmetic or geometric with different examples?
3. How does the value of $r$ affect the growth of a geometric sequence?
4. Can a sequence be both arithmetic and geometric? Why or why not?
5. What are some real-world applications of geometric sequences?

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Tip:

Always test both arithmetic and geometric properties systematically when the sequence type is not specified.