If the altitudes of a triangle are in A.P,then the sides of the triangle are in

Options

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

Answer

To solve the problem, we need to establish the relationship between the altitudes of a triangle and its sides when the altitudes are in Arithmetic Progression (AP).

### Step-by-Step Solution:

1. **Understanding the Problem**:
- Let the altitudes of triangle ABC be denoted as \( h_a, h_b, h_c \) corresponding to sides \( a, b, c \) respectively.
- We are given that \( h_a, h_b, h_c \) are in AP.

2. **Setting Up the Relationship**:
- Since the altitudes are in AP, we can express this as:
\[ 2h_b = h_a + h_c \]

3. **Area of the Triangle**:
- The area \( A \) of triangle ABC can be expressed using the base and height:
\[ A = \frac{1}{2} \times a \times h_a = \frac{1}{2} \times b \times h_b = \frac{1}{2} \times c \times h_c \]
- From this, we can derive:
\[ h_a = \frac{2A}{a}, \quad h_b = \frac{2A}{b}, \quad h_c = \frac{2A}{c} \]

4. **Substituting into the AP Condition**:
- Substitute \( h_a, h_b, h_c \) into the AP condition:
\[ 2 \left(\frac{2A}{b}\right) = \frac{2A}{a} + \frac{2A}{c} \]
- Simplifying this gives:
\[ \frac{4A}{b} = \frac{2A}{a} + \frac{2A}{c} \]

5. **Dividing by \( 2A \)** (assuming \( A \neq 0 \)):
- This simplifies to:
\[ \frac{2}{b} = \frac{1}{a} + \frac{1}{c} \]

6. **Finding a Common Denominator**:
- Rearranging gives:
\[ \frac{2}{b} = \frac{c + a}{ac} \]
- Cross-multiplying yields:
\[ 2ac = b(a + c) \]

7. **Conclusion**:
- The relationship \( 2ac = b(a + c) \) indicates that the sides \( a, b, c \) are in Harmonic Progression (HP).
- Therefore, if the altitudes of a triangle are in AP, the sides of the triangle are in HP.

### Final Answer:
The sides of the triangle are in Harmonic Progression (HP).

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