Suppose `a,b,c gt 1 and n in N, n ge 2`. If `log_(a)(n),log_(b)(n),log_(c )(n)` are in A.P., then `log_(a)b+log_(c )b`=

Answer

To solve the problem, we need to find the value of \( \log_b a + \log_b c \) given that \( \log_a n, \log_b n, \log_c n \) are in arithmetic progression (A.P.).

### Step-by-step Solution:

1. **Understanding the A.P. Condition**:
Since \( \log_a n, \log_b n, \log_c n \) are in A.P., we can use the property of A.P. which states that the middle term is the average of the other two terms. Therefore, we can write:
\[ 2 \log_b n = \log_a n + \log_c n \]

2. **Using Change of Base Formula**:
We can apply the change of base formula for logarithms, which states that:
\[ \log_a n = \frac{\log n}{\log a}, \quad \log_b n = \frac{\log n}{\log b}, \quad \log_c n = \frac{\log n}{\log c} \]
Substituting these into our equation gives:
\[ 2 \cdot \frac{\log n}{\log b} = \frac{\log n}{\log a} + \frac{\log n}{\log c} \]

3. **Simplifying the Equation**:
Since \( \log n \) is common in all terms and is non-zero (as \( n > 1 \)), we can cancel \( \log n \) from both sides:
\[ 2 \cdot \frac{1}{\log b} = \frac{1}{\log a} + \frac{1}{\log c} \]

4. **Rearranging the Equation**:
We can rewrite the equation as:
\[ \frac{2}{\log b} = \frac{1}{\log a} + \frac{1}{\log c} \]

5. **Finding \( \log_b a + \log_b c \)**:
Using the property of logarithms, we know:
\[ \log_b a = \frac{1}{\log a / \log b} \quad \text{and} \quad \log_b c = \frac{1}{\log c / \log b} \]
Thus, we can express \( \log_b a + \log_b c \) as:
\[ \log_b a + \log_b c = \frac{1}{\log a / \log b} + \frac{1}{\log c / \log b} = \frac{\log b}{\log a} + \frac{\log b}{\log c} = \log b \left( \frac{1}{\log a} + \frac{1}{\log c} \right) \]
From the earlier equation, we have:
\[ \frac{1}{\log a} + \frac{1}{\log c} = \frac{2}{\log b} \]
Therefore:
\[ \log_b a + \log_b c = \log b \cdot \frac{2}{\log b} = 2 \]

### Final Answer:
Thus, we conclude that:
\[ \log_b a + \log_b c = 2 \]