If a, b, c are in G.P., then `log_(a) 10, log_(b) 10, log_(c) 10` are in

Options

A. A.P.
B. G.P.
C. H.P.
D. none

Answer

To solve the problem, we need to show that if \( a, b, c \) are in geometric progression (G.P.), then \( \log_{a} 10, \log_{b} 10, \log_{c} 10 \) are in harmonic progression (H.P.).

### Step-by-Step Solution:

1. **Understanding G.P.**:
If \( a, b, c \) are in G.P., then by definition, we have:
\[ b^2 = ac \]

2. **Taking Logarithms**:
We will take logarithms (base 10) of both sides of the equation \( b^2 = ac \):
\[ \log_{10}(b^2) = \log_{10}(ac) \]

3. **Applying Logarithmic Properties**:
Using the property of logarithms that states \( \log_{10}(xy) = \log_{10}(x) + \log_{10}(y) \) and \( \log_{10}(x^n) = n \cdot \log_{10}(x) \), we can rewrite the equation:
\[ 2 \log_{10}(b) = \log_{10}(a) + \log_{10}(c) \]

4. **Rearranging the Equation**:
Rearranging the equation gives us:
\[ 2 \log_{10}(b) - \log_{10}(a) - \log_{10}(c) = 0 \]

5. **Expressing in Terms of Reciprocals**:
We can express this in terms of reciprocals:
\[ \frac{1}{\log_{10}(a)} + \frac{1}{\log_{10}(c)} = \frac{2}{\log_{10}(b)} \]
This indicates that \( \log_{10}(a), \log_{10}(b), \log_{10}(c) \) are in harmonic progression.

### Conclusion:
Thus, we conclude that if \( a, b, c \) are in G.P., then \( \log_{a} 10, \log_{b} 10, \log_{c} 10 \) are in harmonic progression.