In `Delta ABC`, I is the incentre, Area of `DeltaIBC, DeltaIAC and DeltaIAB` are, respectively, `Delta_(1), Delta_(2) and Delta_(3)`. If the values of `Delta_(1), Delta_(2) and Delta_(3)` are in A.P., then the altitudes of the `DeltaABC` are in

Options

A. A.P.
B. G.P.
C. H.P.
D. none of these

Explanation

To solve the problem, we need to analyze the areas of triangles formed by the incenter and the vertices of triangle ABC. We are given that the areas of triangles IBC, IAC, and IAB (denoted as Δ1, Δ2, and Δ3) are in Arithmetic Progression (AP). We need to find the relationship between the altitudes of triangle ABC.

### Step-by-Step Solution:

1. **Understanding the Areas**:
- The area of triangle IBC is Δ1.
- The area of triangle IAC is Δ2.
- The area of triangle IAB is Δ3.
- We know that Δ1, Δ2, and Δ3 are in AP.

2. **Using the Area Formula**:
- The area of triangle IBC can be expressed as:
\[ \Delta_1 = \frac{1}{2} r \cdot a \]
- The area of triangle IAC can be expressed as:
\[ \Delta_2 = \frac{1}{2} r \cdot b \]
- The area of triangle IAB can be expressed as:
\[ \Delta_3 = \frac{1}{2} r \cdot c \]
where \( r \) is the inradius, and \( a, b, c \) are the lengths of the sides opposite to vertices A, B, and C respectively.

3. **Setting Up the AP Condition**:
- Since Δ1, Δ2, and Δ3 are in AP, we can write:
\[ 2\Delta_2 = \Delta_1 + \Delta_3 \]
- Substituting the expressions for Δ1, Δ2, and Δ3:
\[ 2 \left( \frac{1}{2} r \cdot b \right) = \frac{1}{2} r \cdot a + \frac{1}{2} r \cdot c \]
- Simplifying this gives:
\[ r \cdot b = \frac{1}{2} r \cdot (a + c) \]
- Dividing by \( r \) (assuming \( r \neq 0 \)):
\[ b = \frac{1}{2} (a + c) \]

4. **Finding the Altitudes**:
- Let the area of triangle ABC be Δ. The altitudes corresponding to sides \( a, b, c \) can be expressed as:
\[ h_a = \frac{2\Delta}{a}, \quad h_b = \frac{2\Delta}{b}, \quad h_c = \frac{2\Delta}{c} \]
- We need to analyze the relationship between \( h_a, h_b, h_c \).

5. **Relationship Between Altitudes**:
- From the previous step, we can express the altitudes in terms of the area Δ:
\[ h_a = \frac{2\Delta}{a}, \quad h_b = \frac{2\Delta}{\frac{1}{2}(a+c)}, \quad h_c = \frac{2\Delta}{c} \]
- Since \( b = \frac{1}{2}(a + c) \), we can see that:
\[ h_b = \frac{4\Delta}{a + c} \]

6. **Conclusion**:
- The altitudes \( h_a, h_b, h_c \) are related such that if \( \Delta_1, \Delta_2, \Delta_3 \) are in AP, then the altitudes \( h_a, h_b, h_c \) are in Harmonic Progression (HP).

### Final Answer:
The altitudes of triangle ABC are in Harmonic Progression (HP).