If x ∈ R and the numbers `(5 ^ (1−x) +5^ (x+1) , a/2, (25^ x +25^ −x ))` form an A. P. then a must lie in the interval ............ .

Explanation

To solve the problem, we need to determine the interval in which \( a \) lies, given that the numbers \( (5^{(1-x)} + 5^{(x+1)}, \frac{a}{2}, (25^x + 25^{-x})) \) form an arithmetic progression (A.P.).

### Step-by-Step Solution:

1. **Understanding the A.P. Condition**:
For three numbers \( A, B, C \) to be in A.P., the condition is:
\[ 2B = A + C \]
Here, let:
\[ A = 5^{(1-x)} + 5^{(x+1)}, \quad B = \frac{a}{2}, \quad C = 25^x + 25^{-x} \]

2. **Expressing \( C \) in terms of \( 5^x \)**:
We know that \( 25^x = (5^2)^x = (5^x)^2 \) and \( 25^{-x} = (5^{-x})^2 \). Thus:
\[ C = 25^x + 25^{-x} = (5^x)^2 + (5^{-x})^2 \]

3. **Substituting \( t = 5^x \)**:
Let \( t = 5^x \). Then \( 5^{-x} = \frac{1}{t} \). Now, we can rewrite \( A \) and \( C \):
\[ A = 5^{(1-x)} + 5^{(x+1)} = 5 \cdot 5^{-x} + 5^x \cdot 5 = 5 \left(\frac{1}{t} + t\right) \]
\[ C = t^2 + \frac{1}{t^2} \]

4. **Setting up the equation**:
Now substituting \( A \), \( B \), and \( C \) into the A.P. condition:
\[ 2 \cdot \frac{a}{2} = 5 \left(\frac{1}{t} + t\right) + \left(t^2 + \frac{1}{t^2}\right) \]
Simplifying gives:
\[ a = 5 \left(\frac{1}{t} + t\right) + t^2 + \frac{1}{t^2} \]

5. **Simplifying the expression**:
We can further simplify \( a \):
\[ a = 5 \left(\frac{1}{t} + t\right) + t^2 + \frac{1}{t^2} = 5 \left(t + \frac{1}{t}\right) + \left(t^2 + \frac{1}{t^2}\right) \]

6. **Using the inequality**:
We know from the AM-GM inequality that:
\[ t + \frac{1}{t} \geq 2 \quad \text{and} \quad t^2 + \frac{1}{t^2} \geq 2 \]
Therefore:
\[ a \geq 5(2) + 2 = 10 + 2 = 12 \]

7. **Finding the interval for \( a \)**:
Since \( t + \frac{1}{t} \) can take values from \( 2 \) to \( \infty \), \( a \) can take values from \( 12 \) to \( \infty \). Thus, we conclude:
\[ a \in [12, \infty) \]

### Final Answer:
The value of \( a \) must lie in the interval \( [12, \infty) \).