No. of terms in the series `4,2,1,…,(1)/(128)` is :

Options

A. 10
B. 12
C. 14
D. None of these

Explanation

To find the number of terms in the series \(4, 2, 1, \ldots, \frac{1}{128}\), we can identify that this is a geometric progression (GP).

### Step-by-step Solution:

1. **Identify the first term (A) and the common ratio (R)**:
- The first term \(A = 4\).
- The second term is \(2\), so the common ratio \(R\) can be calculated as:
\[ R = \frac{\text{Second term}}{\text{First term}} = \frac{2}{4} = \frac{1}{2} \]

2. **Identify the last term (AN)**:
- The last term of the series is given as \(AN = \frac{1}{128}\).

3. **Use the formula for the nth term of a GP**:
- The formula for the nth term of a geometric progression is:
\[ AN = A \cdot R^{n-1} \]
- Substituting the known values:
\[ \frac{1}{128} = 4 \cdot \left(\frac{1}{2}\right)^{n-1} \]

4. **Rearranging the equation**:
- We can rewrite the equation to isolate the term involving \(n\):
\[ \left(\frac{1}{2}\right)^{n-1} = \frac{1}{128} \cdot \frac{1}{4} \]
- Simplifying the right-hand side:
\[ \frac{1}{128} \cdot \frac{1}{4} = \frac{1}{512} \]

5. **Express \(512\) as a power of \(2\)**:
- We know that \(512 = 2^9\), hence:
\[ \frac{1}{512} = \frac{1}{2^9} = 2^{-9} \]

6. **Set the powers equal**:
- Since the bases are the same, we can equate the exponents:
\[ n - 1 = 9 \]

7. **Solve for \(n\)**:
- Adding \(1\) to both sides gives:
\[ n = 9 + 1 = 10 \]

### Final Answer:
The number of terms in the series is \(10\).