Which term of the sequence `(1)/(2) , (1)/(4) , (1)/(8) , -(1)/(16) ….` is `- (1)/(256)` ?

Options

A. 9 th
B. 8 th
C. 7 th
D. 5 th

Answer

To determine which term of the sequence \( \frac{1}{2}, \frac{1}{4}, \frac{1}{8}, -\frac{1}{16}, \ldots \) is equal to \( -\frac{1}{256} \), we can analyze the pattern of the sequence.

### Step-by-Step Solution:

1. **Identify the pattern in the sequence:**
The sequence starts with positive fractions and then transitions to a negative fraction. The denominators are powers of 2:
- The first term is \( \frac{1}{2} = \frac{1}{2^1} \)
- The second term is \( \frac{1}{4} = \frac{1}{2^2} \)
- The third term is \( \frac{1}{8} = \frac{1}{2^3} \)
- The fourth term is \( -\frac{1}{16} = -\frac{1}{2^4} \)

2. **Generalize the \( n^{th} \) term:**
From the pattern, we can see that the \( n^{th} \) term can be expressed as:
\[ a_n = \frac{(-1)^{n+1}}{2^n} \]
This means that for odd \( n \), the term is positive, and for even \( n \), the term is negative.

3. **Set up the equation:**
We need to find \( n \) such that:
\[ a_n = -\frac{1}{256} \]
Substituting our general formula:
\[ \frac{(-1)^{n+1}}{2^n} = -\frac{1}{256} \]

4. **Solve for \( n \):**
Since \( -\frac{1}{256} \) indicates that \( n \) must be even (because \( (-1)^{n+1} \) must be negative), we can set:
\[ \frac{1}{2^n} = \frac{1}{256} \]
This implies:
\[ 2^n = 256 \]
We know that \( 256 = 2^8 \), so:
\[ n = 8 \]

5. **Conclusion:**
Therefore, the term \( -\frac{1}{256} \) is the 8th term of the sequence.

### Final Answer:
The term \( -\frac{1}{256} \) is the 8th term of the sequence.