If a, ,b c, are in G.P., then `log_(a) n, log_(b) n, log_(c) n `are in

Options

A. A.P.
B. G.P.
C. H.P.
D. None of these

Answer

To solve the problem, we need to show that if \( a, b, c \) are in geometric progression (G.P.), then \( \log_a n, \log_b n, \log_c n \) are in harmonic progression (H.P.).

### Step-by-Step Solution:

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

2. **Expressing Logarithms**:
Let:
\[ x = \log_a n, \quad y = \log_b n, \quad z = \log_c n \]
We can express \( a, b, c \) in terms of \( n \):
\[ a = n^{1/x}, \quad b = n^{1/y}, \quad c = n^{1/z} \]

3. **Substituting into G.P. Condition**:
Using the G.P. condition \( b^2 = ac \):
\[ (n^{1/y})^2 = n^{1/x} \cdot n^{1/z} \]
This simplifies to:
\[ n^{2/y} = n^{1/x + 1/z} \]

4. **Equating Exponents**:
Since the bases are the same, we can equate the exponents:
\[ \frac{2}{y} = \frac{1}{x} + \frac{1}{z} \]

5. **Finding a Common Denominator**:
Rewriting the right-hand side:
\[ \frac{2}{y} = \frac{z + x}{xz} \]

6. **Cross Multiplying**:
Cross multiplying gives:
\[ 2xz = y(x + z) \]

7. **Rearranging**:
Rearranging the equation leads to:
\[ \frac{1}{x}, \frac{1}{y}, \frac{1}{z} \text{ are in H.P.} \]

8. **Conclusion**:
Since \( \frac{1}{x}, \frac{1}{y}, \frac{1}{z} \) are in H.P., it follows that \( \log_a n, \log_b n, \log_c n \) are in H.P.

### Final Answer:
Thus, \( \log_a n, \log_b n, \log_c n \) are in harmonic progression (H.P.).

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