The least value of a for which `5^(1+x)+5^(1-x),a//2,25^(x)+25^(-x)` are three consecutive terms of an A.P., is

Options

A. 10
B. 5
C. 12
D. none of these

Explanation

To find the least value of \( a \) for which \( 5^{1+x} + 5^{1-x}, \frac{a}{2}, 25^x + 25^{-x} \) are three consecutive terms of an arithmetic progression (A.P.), we can follow these steps:

### Step 1: Set up the condition for A.P.
For three terms \( A, B, C \) to be in A.P., the condition is:
\[ 2B = A + C \]
In our case:
- \( A = 5^{1+x} + 5^{1-x} \)
- \( B = \frac{a}{2} \)
- \( C = 25^x + 25^{-x} \)

Substituting these into the A.P. condition gives:
\[ 2 \cdot \frac{a}{2} = (5^{1+x} + 5^{1-x}) + (25^x + 25^{-x}) \]
This simplifies to:
\[ a = 5^{1+x} + 5^{1-x} + 25^x + 25^{-x} \]

### Step 2: Simplify the terms
We can rewrite \( 25^x \) and \( 25^{-x} \) in terms of base 5:
\[ 25^x = (5^2)^x = 5^{2x} \quad \text{and} \quad 25^{-x} = (5^2)^{-x} = 5^{-2x} \]
Thus, we have:
\[ a = 5^{1+x} + 5^{1-x} + 5^{2x} + 5^{-2x} \]

### Step 3: Combine like terms
Now, we can combine the terms:
\[ a = 5^{1+x} + 5^{1-x} + 5^{2x} + 5^{-2x} \]

### Step 4: Find the minimum value of \( a \)
To find the minimum value of \( a \), we can analyze the terms:
- \( 5^{1+x} + 5^{1-x} \) can be expressed as \( 5 \left( 5^x + 5^{-x} \right) \)
- \( 5^{2x} + 5^{-2x} \) can be expressed as \( 5^{2x} + 5^{-2x} \)

Using the property of exponential functions, we know that \( 5^x + 5^{-x} \geq 2 \) (by AM-GM inequality), and similarly for \( 5^{2x} + 5^{-2x} \):
\[ 5^{2x} + 5^{-2x} \geq 2 \]

### Step 5: Substitute minimum values
Substituting these minimum values into the expression for \( a \):
\[ a \geq 5 \cdot 2 + 2 = 10 + 2 = 12 \]

### Conclusion
Thus, the least value of \( a \) for which the terms are in A.P. is:
\[ \boxed{12} \]