If p, q, r are in AP as well as G.P., then which one of the following is correct?

Options

A. `p=qner`
B. `pneqner`
C. `pneq=r`
D. `p=q=r`

Explanation

To solve the problem, we need to analyze the conditions given: \( p, q, r \) are in both Arithmetic Progression (AP) and Geometric Progression (GP).

### Step-by-Step Solution:

1. **Understanding AP Condition**:
- If \( p, q, r \) are in AP, then the condition is:
\[ 2q = p + r \]

2. **Understanding GP Condition**:
- If \( p, q, r \) are in GP, then the condition is:
\[ q^2 = pr \]

3. **Expressing \( q \) in terms of \( p \) and \( r \)**:
- From the AP condition, we can rearrange it to express \( q \):
\[ q = \frac{p + r}{2} \]

4. **Substituting \( q \) into the GP condition**:
- Substitute \( q \) into the GP condition:
\[ \left(\frac{p + r}{2}\right)^2 = pr \]
- Expanding the left side:
\[ \frac{(p + r)^2}{4} = pr \]
- Multiplying both sides by 4 to eliminate the fraction:
\[ (p + r)^2 = 4pr \]

5. **Expanding and rearranging**:
- Expanding the left side:
\[ p^2 + 2pr + r^2 = 4pr \]
- Rearranging gives:
\[ p^2 + r^2 - 2pr = 0 \]

6. **Factoring the equation**:
- The equation can be factored as:
\[ (p - r)^2 = 0 \]
- This implies:
\[ p - r = 0 \quad \Rightarrow \quad p = r \]

7. **Finding \( q \)**:
- Now, substituting \( p = r \) back into the expression for \( q \):
\[ q = \frac{p + r}{2} = \frac{p + p}{2} = \frac{2p}{2} = p \]
- Thus, we have:
\[ q = p \]
- Since \( p = r \), we conclude:
\[ p = q = r \]

### Conclusion:
The correct relationship among \( p, q, r \) is:
\[ p = q = r \]