.`1/(p+q),1/(q+r),1/(r+p)` are in AP, then

Options

A. p, q, r in A.P.
B. `q^2,p^2,r^2` are in A.P.
C. `p^2,q^2,r^2` are in A.P.
D. p,q,r in G.P.

Answer

To solve the problem, we need to show that if \( \frac{1}{p+q}, \frac{1}{q+r}, \frac{1}{r+p} \) are in Arithmetic Progression (AP), then \( p^2, q^2, r^2 \) are also in AP.

### Step-by-step Solution:

1. **Understanding the condition for AP**:
- For three terms \( a, b, c \) to be in AP, the condition is:
\[ 2b = a + c \]
- Here, let \( a = \frac{1}{p+q} \), \( b = \frac{1}{q+r} \), and \( c = \frac{1}{r+p} \).

2. **Setting up the equation**:
- According to the AP condition:
\[ 2 \cdot \frac{1}{q+r} = \frac{1}{p+q} + \frac{1}{r+p} \]

3. **Finding a common denominator**:
- The common denominator for the right side is \( (p+q)(r+p) \):
\[ 2 \cdot \frac{1}{q+r} = \frac{(r+p) + (p+q)}{(p+q)(r+p)} \]
- Simplifying the right side:
\[ 2 \cdot \frac{1}{q+r} = \frac{r + 2p + q}{(p+q)(r+p)} \]

4. **Cross-multiplying**:
- Cross-multiplying gives:
\[ 2(p+q)(r+p) = (q+r)(r + 2p + q) \]

5. **Expanding both sides**:
- Left side:
\[ 2(pr + p^2 + qr + qp) \]
- Right side:
\[ (qr + rq + 2pq + q^2 + r^2) \]

6. **Setting the equation**:
- After expanding, we have:
\[ 2pr + 2p^2 + 2qr + 2qp = qr + rq + 2pq + q^2 + r^2 \]

7. **Rearranging the terms**:
- Move all terms to one side:
\[ 2pr + 2p^2 - qr - rq - 2pq - q^2 - r^2 = 0 \]

8. **Grouping terms**:
- Grouping terms leads to:
\[ 2p^2 - q^2 - r^2 + 2pr - 2pq - qr = 0 \]

9. **Identifying the AP condition**:
- Rearranging gives:
\[ 2p^2 = q^2 + r^2 \]
- This shows that \( p^2, q^2, r^2 \) are in AP since it satisfies the condition \( 2p^2 = q^2 + r^2 \).

### Conclusion:
Thus, we conclude that if \( \frac{1}{p+q}, \frac{1}{q+r}, \frac{1}{r+p} \) are in AP, then \( p^2, q^2, r^2 \) are also in AP.