If the sides of a triangle are in A.P. as well as in G.P., then the value of `r_1/r_2 - r_2/r_3` is

Options

A. 1
B. 0
C. 2r
D. none

Answer

To solve the problem, we need to find the value of \( \frac{r_1}{r_2} - \frac{r_2}{r_3} \) given that the sides of a triangle are in both Arithmetic Progression (A.P.) and Geometric Progression (G.P.).

### Step 1: Understanding the sides in A.P. and G.P.
Let the sides of the triangle be \( a, b, c \).
- Since they are in A.P., we can express them as:
\[ b = \frac{a + c}{2} \]
- Since they are also in G.P., we have:
\[ b^2 = ac \]

### Step 2: Setting up the equations
From the A.P. condition, we can rearrange it to:
\[ 2b = a + c \quad \text{(1)} \]
From the G.P. condition, we can write:
\[ b^2 = ac \quad \text{(2)} \]

### Step 3: Solving the equations
From equation (1), we can express \( a \) and \( c \) in terms of \( b \):
\[ a + c = 2b \implies c = 2b - a \quad \text{(3)} \]
Substituting equation (3) into equation (2):
\[ b^2 = a(2b - a) \]
Expanding this gives:
\[ b^2 = 2ab - a^2 \]
Rearranging gives us a quadratic equation:
\[ a^2 - 2ab + b^2 = 0 \]

### Step 4: Solving the quadratic equation
Using the quadratic formula \( a = \frac{-B \pm \sqrt{B^2 - 4AC}}{2A} \):
Here, \( A = 1, B = -2b, C = b^2 \):
\[ a = \frac{2b \pm \sqrt{(-2b)^2 - 4 \cdot 1 \cdot b^2}}{2 \cdot 1} \]
\[ a = \frac{2b \pm \sqrt{4b^2 - 4b^2}}{2} \]
\[ a = \frac{2b \pm 0}{2} = b \]
Thus, \( a = b \) and substituting back into equation (3):
\[ c = 2b - b = b \implies a = b = c \]

### Step 5: Finding the circumradius and inradius
For an equilateral triangle (since \( a = b = c \)):
- The semi-perimeter \( s \) is:
\[ s = \frac{a + b + c}{2} = \frac{3b}{2} \]
- The area \( \Delta \) of the triangle is:
\[ \Delta = \frac{\sqrt{3}}{4} a^2 = \frac{\sqrt{3}}{4} b^2 \]
- The circumradius \( R \) is given by:
\[ R = \frac{abc}{4\Delta} = \frac{b^3}{4 \cdot \frac{\sqrt{3}}{4} b^2} = \frac{b}{\sqrt{3}} \]
- The inradius \( r \) is given by:
\[ r = \frac{\Delta}{s} = \frac{\frac{\sqrt{3}}{4} b^2}{\frac{3b}{2}} = \frac{\sqrt{3}b}{6} \]

### Step 6: Calculating \( \frac{r_1}{r_2} - \frac{r_2}{r_3} \)
Since \( r_1 = R, r_2 = r, r_3 = r \) (as all sides are equal):
\[ \frac{r_1}{r_2} = \frac{\frac{b}{\sqrt{3}}}{\frac{\sqrt{3}b}{6}} = \frac{6}{3} = 2 \]
\[ \frac{r_2}{r_3} = \frac{r}{r} = 1 \]
Thus:
\[ \frac{r_1}{r_2} - \frac{r_2}{r_3} = 2 - 1 = 1 \]

### Final Answer:
The value of \( \frac{r_1}{r_2} - \frac{r_2}{r_3} \) is \( 1 \).