If `a, b, c` are in `G.P. (a, b, c epsilon R')` then `(1)/(1+loga), (1)/(1+logb), (1)/(1+logc)` are in

Options

A. `A.P`
B. `G.p`
C. `H.P`
D. `A.G.P.`

Explanation

To solve the problem, we need to show that if \( a, b, c \) are in Geometric Progression (G.P.), then \( \frac{1}{1+\log a}, \frac{1}{1+\log b}, \frac{1}{1+\log c} \) are in Harmonic Progression (H.P.).

### Step-by-Step Solution:

1. **Understanding G.P.**:
Since \( a, b, c \) are in G.P., we can express this relationship as:
\[ b^2 = ac \]

2. **Taking Logarithms**:
Taking the logarithm of both sides gives:
\[ \log b^2 = \log(ac) \]
This simplifies to:
\[ 2 \log b = \log a + \log c \]

3. **Rearranging the Logarithmic Equation**:
We can rewrite the equation as:
\[ \log a + \log c = 2 \log b \]
This indicates that \( \log a, \log b, \log c \) are in Arithmetic Progression (A.P.).

4. **Adding a Constant**:
If \( \log a, \log b, \log c \) are in A.P., then adding the same constant (which is 1 in this case) to each term gives:
\[ 1 + \log a, 1 + \log b, 1 + \log c \]
These terms will still maintain the property of being in A.P.

5. **Reciprocal of A.P.**:
The reciprocals of terms in A.P. are in H.P. Therefore, we take the reciprocals:
\[ \frac{1}{1 + \log a}, \frac{1}{1 + \log b}, \frac{1}{1 + \log c} \]
Since \( 1 + \log a, 1 + \log b, 1 + \log c \) are in A.P., their reciprocals are in H.P.

6. **Conclusion**:
Thus, we conclude that:
\[ \frac{1}{1 + \log a}, \frac{1}{1 + \log b}, \frac{1}{1 + \log c} \text{ are in H.P.} \]

### Final Answer:
The terms \( \frac{1}{1+\log a}, \frac{1}{1+\log b}, \frac{1}{1+\log c} \) are in Harmonic Progression (H.P.).