In an equilateral `DeltaABC`, (where symbols used have usual meanings), then r, R and `r_(1)` form :
i) an A.P.
ii)a G.P.
iii) an H.P.
iv) neither an A.P., G.P. nor H.P.

Options

A. an A.P.
B. a G.P.
C. an H.P.
D. neither an A.P., G.P. nor H.P.

Answer

To solve the problem, we need to find the relationships between the inradius (r), circumradius (R), and the exradius (R1) of an equilateral triangle. Let's denote the side length of the equilateral triangle as \( L \).

### Step 1: Calculate the Inradius (r)
The formula for the inradius \( r \) of a triangle is given by:
\[ r = \frac{\Delta}{s} \]
where \( \Delta \) is the area of the triangle and \( s \) is the semi-perimeter.

For an equilateral triangle:
- The area \( \Delta \) is given by:
\[ \Delta = \frac{\sqrt{3}}{4} L^2 \]
- The semi-perimeter \( s \) is:
\[ s = \frac{3L}{2} \]

Now substituting these values into the formula for \( r \):
\[ r = \frac{\frac{\sqrt{3}}{4} L^2}{\frac{3L}{2}} = \frac{\sqrt{3} L^2}{4} \cdot \frac{2}{3L} = \frac{\sqrt{3} L}{6} \]

### Step 2: Calculate the Circumradius (R)
The formula for the circumradius \( R \) of a triangle is given by:
\[ R = \frac{abc}{4\Delta} \]
For an equilateral triangle, \( a = b = c = L \):
\[ R = \frac{L \cdot L \cdot L}{4 \cdot \frac{\sqrt{3}}{4} L^2} = \frac{L^3}{\sqrt{3} L^2} = \frac{L}{\sqrt{3}} \]

### Step 3: Calculate the Exradius (R1)
The exradius \( R1 \) opposite to side \( a \) is given by:
\[ R1 = \frac{\Delta}{s - a} \]
Substituting the values:
\[ R1 = \frac{\frac{\sqrt{3}}{4} L^2}{\frac{3L}{2} - L} = \frac{\frac{\sqrt{3}}{4} L^2}{\frac{L}{2}} = \frac{\sqrt{3} L^2}{4} \cdot \frac{2}{L} = \frac{\sqrt{3} L}{2} \]

### Step 4: Establish the relationship between r, R, and R1
Now we have:
- \( r = \frac{\sqrt{3} L}{6} \)
- \( R = \frac{L}{\sqrt{3}} \)
- \( R1 = \frac{\sqrt{3} L}{2} \)

Now, let's express these in a common format:
- \( r = \frac{\sqrt{3} L}{6} \)
- \( R = \frac{2\sqrt{3} L}{6} \)
- \( R1 = \frac{3\sqrt{3} L}{6} \)

### Step 5: Check if they form an A.P.
To check if \( r \), \( R \), and \( R1 \) form an Arithmetic Progression (A.P.), we need to verify if:
\[ 2R = r + R1 \]
Substituting the values:
\[ 2 \cdot \frac{2\sqrt{3} L}{6} = \frac{\sqrt{3} L}{6} + \frac{3\sqrt{3} L}{6} \]
This simplifies to:
\[ \frac{4\sqrt{3} L}{6} = \frac{4\sqrt{3} L}{6} \]
This equality holds true, confirming that \( r \), \( R \), and \( R1 \) indeed form an A.P.

### Final Answer
Thus, the correct option is:
**i) an A.P.**