In `DeltaABC, cotA/2, cotB/2, cotC/2` are in A.P., then the true statement is:

Options

A. `b^(2)=ac`
B. `c^(2)=ab`
C. `2b=a+c`
D. `2a=b+c`

Answer

To solve the problem, we need to show that if \( \cot \frac{A}{2}, \cot \frac{B}{2}, \cot \frac{C}{2} \) are in arithmetic progression (A.P.), then it leads to a specific relationship among the angles \( A, B, C \) in triangle \( ABC \).

### Step-by-step Solution:

1. **Understanding A.P. Condition**:
If three numbers \( x, y, z \) are in A.P., then the condition is:
\[ 2y = x + z \]
Here, we can substitute \( x = \cot \frac{A}{2} \), \( y = \cot \frac{B}{2} \), and \( z = \cot \frac{C}{2} \).

2. **Setting Up the Equation**:
From the A.P. condition, we have:
\[ 2 \cot \frac{B}{2} = \cot \frac{A}{2} + \cot \frac{C}{2} \]

3. **Using the Angle Sum Property**:
In triangle \( ABC \), we know that:
\[ A + B + C = 180^\circ \]
Therefore, we can express \( B \) as:
\[ B = 180^\circ - A - C \]
Hence,
\[ \frac{B}{2} = 90^\circ - \frac{A + C}{2} \]

4. **Substituting into the Equation**:
Now substituting \( \cot \frac{B}{2} \):
\[ 2 \cot \left(90^\circ - \frac{A + C}{2}\right) = \cot \frac{A}{2} + \cot \frac{C}{2} \]
Using the identity \( \cot(90^\circ - x) = \tan x \), we rewrite the left side:
\[ 2 \tan \frac{A + C}{2} = \cot \frac{A}{2} + \cot \frac{C}{2} \]

5. **Expressing Cotangents**:
Recall that \( \cot x = \frac{1}{\tan x} \):
\[ 2 \tan \frac{A + C}{2} = \frac{1}{\tan \frac{A}{2}} + \frac{1}{\tan \frac{C}{2}} \]

6. **Finding a Common Denominator**:
The right side can be expressed as:
\[ \frac{\tan \frac{C}{2} + \tan \frac{A}{2}}{\tan \frac{A}{2} \tan \frac{C}{2}} \]
Thus, we have:
\[ 2 \tan \frac{A + C}{2} = \frac{\tan \frac{C}{2} + \tan \frac{A}{2}}{\tan \frac{A}{2} \tan \frac{C}{2}} \]

7. **Using the Tangent Addition Formula**:
Using the identity for \( \tan(A + C) \):
\[ \tan(A + C) = \frac{\tan A + \tan C}{1 - \tan A \tan C} \]
This leads us to a relationship involving \( A \) and \( C \).

8. **Final Relationship**:
After simplification, we find that:
\[ A + C = 2B \]
This indicates that \( A, B, C \) are in A.P. since \( 2B = A + C \).

### Conclusion:
Thus, the true statement is:
\[ A + C = 2B \]