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

Options

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

Answer

To solve the problem, we need to show that if the radii \( r_1, r_2, r_3 \) are in Harmonic Progression (H.P.), then the sides of the triangle corresponding to these radii are in Arithmetic Progression (A.P.).

### Step-by-Step Solution:

1. **Understanding the Relationship**:
- The radii \( r_1, r_2, r_3 \) are given to be in H.P. This means that the reciprocals \( \frac{1}{r_1}, \frac{1}{r_2}, \frac{1}{r_3} \) are in A.P.
- The condition for three numbers \( x, y, z \) to be in A.P. is given by \( 2y = x + z \).

2. **Using the Formula for the Radii**:
- The radius \( r_1 \) of the incircle is given by:
\[ r_1 = \frac{\Delta}{s - a} \]
- The radius \( r_2 \) of the circle opposite side \( b \) is:
\[ r_2 = \frac{\Delta}{s - b} \]
- The radius \( r_3 \) of the circle opposite side \( c \) is:
\[ r_3 = \frac{\Delta}{s - c} \]
- Here, \( \Delta \) is the area of the triangle and \( s \) is the semi-perimeter given by \( s = \frac{a + b + c}{2} \).

3. **Setting Up the H.P. Condition**:
- Since \( r_1, r_2, r_3 \) are in H.P., we have:
\[ \frac{2}{r_2} = \frac{1}{r_1} + \frac{1}{r_3} \]

4. **Substituting the Formulas**:
- Substitute the expressions for \( r_1, r_2, r_3 \):
\[ \frac{2}{\frac{\Delta}{s - b}} = \frac{1}{\frac{\Delta}{s - a}} + \frac{1}{\frac{\Delta}{s - c}} \]
- This simplifies to:
\[ \frac{2(s - b)}{\Delta} = \frac{s - a}{\Delta} + \frac{s - c}{\Delta} \]
- Multiplying through by \( \Delta \) (assuming \( \Delta \neq 0 \)):
\[ 2(s - b) = (s - a) + (s - c) \]

5. **Simplifying the Equation**:
- Rearranging the equation gives:
\[ 2s - 2b = 2s - a - c \]
- Canceling \( 2s \) from both sides results in:
\[ 2b = a + c \]

6. **Conclusion**:
- The equation \( 2b = a + c \) shows that the sides \( a, b, c \) are in Arithmetic Progression (A.P.).

### Final Result:
Thus, if \( r_1, r_2, r_3 \) are in H.P., then the sides \( a, b, c \) of the triangle are in A.P.

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