If `5^(1 + x) + 5^(1 - x), (a)/(2), 25^(x) + 25^(-x)` are three consecutive terms of an A.P., then which of the following is true ?

Options

A. `a le 12`
B. `a lt 12`
C. `a ge 12`
D. `a gt 12`

Answer

To solve the problem, we need to establish that the three expressions \(5^{(1+x)} + 5^{(1-x)}\), \(\frac{a}{2}\), and \(25^{x} + 25^{-x}\) are in Arithmetic Progression (A.P.).

### Step-by-step Solution:

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

2. **Substituting the Terms**:
We substitute \(a\) and \(c\) into the A.P. condition:
\[ 2 \left(\frac{a}{2}\right) = (5^{(1+x)} + 5^{(1-x)}) + (25^{x} + 25^{-x}) \]
This simplifies to:
\[ a = 5^{(1+x)} + 5^{(1-x)} + 25^{x} + 25^{-x} \]

3. **Expressing \(25^{x}\) in terms of \(5^{x}\)**:
Since \(25 = 5^2\), we can rewrite \(25^{x}\) and \(25^{-x}\):
\[ 25^{x} = (5^2)^{x} = 5^{2x}, \quad 25^{-x} = (5^2)^{-x} = 5^{-2x} \]
Thus, we have:
\[ c = 5^{2x} + 5^{-2x} \]

4. **Combining the Terms**:
Now, we can express \(a\) as:
\[ a = 5^{(1+x)} + 5^{(1-x)} + 5^{2x} + 5^{-2x} \]
This can be simplified further:
\[ a = 5 \cdot 5^{x} + \frac{5}{5^{x}} + 5^{2x} + \frac{1}{5^{2x}} \]

5. **Using the A.P. Condition**:
Now, we need to check if:
\[ 5^{(1+x)} + 5^{(1-x)} + 5^{2x} + 5^{-2x} = 2 \cdot \frac{a}{2} \]
This leads to:
\[ 5^{(1+x)} + 5^{(1-x)} = 5^{2x} + 5^{-2x} \]

6. **Applying AM-GM Inequality**:
We can apply the AM-GM inequality to both sides:
\[ \frac{5^{(1+x)} + 5^{(1-x)}}{2} \geq \sqrt{5^{(1+x)} \cdot 5^{(1-x)}} \]
Simplifying the right-hand side gives:
\[ \sqrt{5^{(1+x)} \cdot 5^{(1-x)}} = \sqrt{5^{1+x+1-x}} = \sqrt{5^2} = 5 \]
Therefore:
\[ 5^{(1+x)} + 5^{(1-x)} \geq 10 \]

7. **Conclusion**:
Since \(5^{(1+x)} + 5^{(1-x)}\) is greater than or equal to 10, we conclude that:
\[ a \geq 10 \]
This implies that the correct option is that \(a\) is greater than or equal to 10.