If `r_1 , r_2 , r_3` in a triangle be in H.P. then the sides are

Options

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

Answer

To solve the problem, we need to understand the relationship between the inradii of a triangle and the sides of the triangle when the inradii are in Harmonic Progression (H.P.).

### Step-by-Step Solution:

1. **Understanding the Inradii**:
The inradii \( r_1, r_2, r_3 \) correspond to the incircles of the triangle formed by the sides opposite to vertices A, B, and C respectively. The inradii can be expressed as:
\[ r_1 = \frac{\Delta}{s - a}, \quad r_2 = \frac{\Delta}{s - b}, \quad r_3 = \frac{\Delta}{s - c} \]
where \( \Delta \) is the area of the triangle and \( s \) is the semi-perimeter given by \( s = \frac{a + b + c}{2} \).

2. **Condition for H.P.**:
For \( r_1, r_2, r_3 \) to be in H.P., the reciprocals \( \frac{1}{r_1}, \frac{1}{r_2}, \frac{1}{r_3} \) must be in A.P. (Arithmetic Progression). This means:
\[ 2 \cdot \frac{1}{r_2} = \frac{1}{r_1} + \frac{1}{r_3} \]

3. **Substituting the Inradii**:
Substituting the expressions for \( r_1, r_2, r_3 \):
\[ 2 \cdot \frac{s - b}{\Delta} = \frac{s - a}{\Delta} + \frac{s - c}{\Delta} \]
Simplifying this, we can eliminate \( \Delta \):
\[ 2(s - b) = (s - a) + (s - c) \]

4. **Rearranging the Equation**:
Rearranging gives us:
\[ 2s - 2b = 2s - a - c \]
Simplifying further, we find:
\[ 2b = a + c \]

5. **Conclusion**:
The relationship \( 2b = a + c \) indicates that the sides of the triangle satisfy the condition of being in an isosceles triangle, where the side opposite to the angle corresponding to \( b \) is equal to the average of the other two sides.

### Final Result:
Thus, if \( r_1, r_2, r_3 \) are in H.P., the sides of the triangle are such that:
\[ 2b = a + c \]
This implies that the triangle is isosceles with respect to the side \( b \).