Which term of the sequence $$ 32 ;-16 ; 8 ; \ldots $$ is $$ \frac{1}{8} ? $$

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To determine which term of the sequence $$ 32, -16, 8, \ldots $$ is $$ \frac{1}{8} $$, let's analyze the sequence step by step.

### Step 1: Identify the Sequence Type
The given sequence is:
- 1st term: $$ 32 $$
- 2nd term: $$ -16 $$
- 3rd term: $$ 8 $$

We can observe that each term is obtained by multiplying the previous term by $$-\frac{1}{2}$$:
- $$ 32 \times \left(-\frac{1}{2}\right) = -16 $$
- $$ -16 \times \left(-\frac{1}{2}\right) = 8 $$

Thus, the sequence is a **geometric sequence** with:
- **First term ($$ a $$)**: $$ 32 $$
- **Common ratio ($$ r $$)**: $$ -\frac{1}{2} $$

### Step 2: Formula for the $$ n $$-th Term
The formula for the $$ n $$-th term of a geometric sequence is:
$$
a_n = a \cdot r^{n-1}
$$
We need to find $$ n $$ such that:
$$
32 \cdot \left(-\frac{1}{2}\right)^{n-1} = \frac{1}{8}
$$

### Step 3: Solve for $$ n $$
First, express both sides with base 2:
$$
32 = 2^5 \quad \text{and} \quad \frac{1}{8} = 2^{-3}
$$
Substitute into the equation:
$$
2^5 \cdot \left(-\frac{1}{2}\right)^{n-1} = 2^{-3}
$$
Simplify the equation by expressing $$-\frac{1}{2}$$ as $$ (-1) \cdot 2^{-1} $$:
$$
2^5 \cdot (-1)^{n-1} \cdot 2^{-(n-1)} = 2^{-3}
$$
Combine the exponents of 2:
$$
2^{5 - (n-1)} \cdot (-1)^{n-1} = 2^{-3}
$$
$$
2^{6 - n} \cdot (-1)^{n-1} = 2^{-3}
$$
For the magnitudes to be equal and the expression to be positive:
$$
6 - n = -3 \quad \Rightarrow \quad n = 9
$$
Additionally, since $$ \frac{1}{8} $$ is positive, $$ (-1)^{n-1} $$ must also be positive. This occurs when $$ n-1 $$ is even, which is true for $$ n = 9 $$.

### Final Answer
The term $$ \frac{1}{8} $$ is the **9th term** of the sequence.

**Answer:** 9
The 9th term of the sequence is $$ \frac{1}{8} $$.

Which term of the sequence
is

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