If r and r' be the maximum ranges up and down the inclined plane respectively and R be max. range on horizontal plane, then r, R, r' are in

Options

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

Answer

To solve the problem, we need to relate the maximum ranges \( r \), \( r' \), and \( R \) based on their definitions and the equations governing projectile motion on inclined planes.

### Step-by-Step Solution:

1. **Define the Variables**:
- Let \( r \) be the maximum range on a horizontal plane.
- Let \( r' \) be the maximum range down the inclined plane.
- Let \( r \) be the maximum range up the inclined plane.

2. **Maximum Range on Horizontal Plane**:
- The maximum range \( R \) on a horizontal plane is given by the formula:
\[ R = \frac{U^2}{g} \]
where \( U \) is the initial velocity and \( g \) is the acceleration due to gravity.

3. **Maximum Range Up the Inclined Plane**:
- The formula for the maximum range \( r \) up the inclined plane at an angle \( \theta \) is:
\[ r = \frac{U^2}{g} \cdot (1 + \sin \theta) \]

4. **Maximum Range Down the Inclined Plane**:
- The formula for the maximum range \( r' \) down the inclined plane is:
\[ r' = \frac{U^2}{g} \cdot (1 - \sin \theta) \]

5. **Relating the Ranges**:
- From the equations for \( r \) and \( r' \), we can express them in terms of \( R \):
\[ r = R(1 + \sin \theta) \]
\[ r' = R(1 - \sin \theta) \]

6. **Combining the Equations**:
- Now, we can relate \( r \) and \( r' \) to \( R \):
\[ \frac{1}{r} + \frac{1}{r'} = \frac{1}{R(1 + \sin \theta)} + \frac{1}{R(1 - \sin \theta)} \]
- Simplifying this gives:
\[ \frac{1}{r} + \frac{1}{r'} = \frac{(1 - \sin \theta) + (1 + \sin \theta)}{R(1 - \sin^2 \theta)} = \frac{2}{R \cos^2 \theta} \]

7. **Final Relation**:
- Therefore, we can conclude that:
\[ \frac{1}{r} + \frac{1}{r'} = \frac{2g}{U^2} \]

### Conclusion:
The relationship between the maximum ranges \( r \), \( r' \), and \( R \) can be summarized as:
\[ \frac{1}{r} + \frac{1}{r'} = \frac{2}{R} \]