`log_3 2, log_6 2, log_12 2` are in

Options

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

Answer

To determine whether \( \log_3 2 \), \( \log_6 2 \), and \( \log_{12} 2 \) are in Arithmetic Progression (AP), Geometric Progression (GP), or Harmonic Progression (HP), we will follow these steps:

### Step 1: Define the logarithmic terms
Let:
- \( A = \log_3 2 \)
- \( B = \log_6 2 \)
- \( C = \log_{12} 2 \)

### Step 2: Convert logarithms to a common base
Using the change of base formula, we can express these logarithms in terms of a common base (base 2):
\[ A = \frac{\log_2 2}{\log_2 3} = \frac{1}{\log_2 3} \]
\[ B = \frac{\log_2 2}{\log_2 6} = \frac{1}{\log_2 6} \]
\[ C = \frac{\log_2 2}{\log_2 12} = \frac{1}{\log_2 12} \]

### Step 3: Find the reciprocals
Now, we take the reciprocals of \( A \), \( B \), and \( C \):
\[ \frac{1}{A} = \log_2 3 \]
\[ \frac{1}{B} = \log_2 6 \]
\[ \frac{1}{C} = \log_2 12 \]

### Step 4: Simplify the logarithms
Using the properties of logarithms:
\[ \log_2 6 = \log_2 (2 \cdot 3) = \log_2 2 + \log_2 3 = 1 + \log_2 3 \]
\[ \log_2 12 = \log_2 (4 \cdot 3) = \log_2 4 + \log_2 3 = 2 + \log_2 3 \]

### Step 5: Check for Arithmetic Progression
Now we check if \( \frac{1}{A} \), \( \frac{1}{B} \), and \( \frac{1}{C} \) are in AP:
- The condition for three numbers \( x, y, z \) to be in AP is \( 2y = x + z \).
- Here, we have:
\[ 2 \log_2 6 = 2(1 + \log_2 3) = 2 + 2\log_2 3 \]
\[ \log_2 3 + \log_2 12 = \log_2 3 + (2 + \log_2 3) = 2 + 2\log_2 3 \]
- Since both sides are equal, \( \frac{1}{A}, \frac{1}{B}, \frac{1}{C} \) are in AP.

### Step 6: Conclusion about the original terms
Since \( \frac{1}{A}, \frac{1}{B}, \frac{1}{C} \) are in AP, it follows that \( A, B, C \) are in Harmonic Progression (HP).

### Final Answer
Thus, \( \log_3 2, \log_6 2, \log_{12} 2 \) are in Harmonic Progression (HP).

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