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 series is a geometric progression (GP) with the first term \(a = 4\) and a common ratio \(r\).

### Step-by-step Solution:

1. **Identify the first term and common ratio:**
- 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. **Write the formula for the nth term of a GP:**
- The formula for the nth term \(a_n\) of a geometric progression is given by:
\[ a_n = a \cdot r^{n-1} \]
- Here, we want to find \(n\) such that \(a_n = \frac{1}{128}\).

3. **Set up the equation:**
- Substitute \(a\) and \(r\) into the formula:
\[ \frac{1}{128} = 4 \cdot \left(\frac{1}{2}\right)^{n-1} \]

4. **Simplify the equation:**
- Divide both sides by 4:
\[ \frac{1}{128} \div 4 = \left(\frac{1}{2}\right)^{n-1} \]
- This simplifies to:
\[ \frac{1}{512} = \left(\frac{1}{2}\right)^{n-1} \]

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

6. **Equate the powers:**
- Now, we can equate the powers since the bases are the same:
\[ -9 = -(n-1) \]
- This simplifies to:
\[ n - 1 = 9 \]

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

### Conclusion:
The number of terms in the series \(4, 2, 1, \ldots, \frac{1}{128}\) is \(10\).

---