If p,q and r are in AP as well as GP, then which one of the following is correct ?

Options

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

Answer

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

### Step-by-Step Solution:

1. **Understanding AP Condition**:
- If p, q, and r are in AP, then the condition is:
\[ 2q = p + r \]
- This means that the middle term (q) is the average of the other two terms (p and r).

2. **Understanding GP Condition**:
- If p, q, and r are in GP, then the condition is:
\[ q^2 = p \cdot r \]
- This means that the square of the middle term (q) is equal to the product of the other two terms (p and r).

3. **Substituting the GP Condition into the AP Condition**:
- From the GP condition, we can express q as:
\[ q = \sqrt{p \cdot r} \]
- Now, substitute this expression for q into the AP condition:
\[ 2(\sqrt{p \cdot r}) = p + r \]

4. **Rearranging the Equation**:
- Rearranging gives us:
\[ p + r - 2\sqrt{p \cdot r} = 0 \]

5. **Recognizing a Perfect Square**:
- We can recognize that the left-hand side can be rewritten as:
\[ (\sqrt{p} - \sqrt{r})^2 = 0 \]
- This implies that:
\[ \sqrt{p} - \sqrt{r} = 0 \quad \Rightarrow \quad \sqrt{p} = \sqrt{r} \]

6. **Conclusion**:
- Squaring both sides gives us:
\[ p = r \]
- Since p = r, and substituting back into the equation for q, we find:
\[ q = p = r \]
- Therefore, the correct conclusion is that all three terms are equal:
\[ p = q = r \]

### Final Answer:
The correct option is that \( p = q = r \).