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

Options

A. p, q, r are in A.P.
B. `p^(2), q^(2), r^(2)` are in A.P.
C. 1/p, 1/q, 1/r are in A.P.
D. none

Answer

To determine the relationship between \( p, q, \) and \( r \) given that the terms \( \frac{1}{q + r}, \frac{1}{r + p}, \frac{1}{p + q} \) are in Arithmetic Progression (A.P.), we can follow these steps:

### Step-by-Step Solution:

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

2. **Setting Up the Equation**:
According to the A.P. condition:
\[ 2 \cdot \frac{1}{r + p} = \frac{1}{q + r} + \frac{1}{p + q} \]

3. **Simplifying the Right Side**:
To combine the right side, we need a common denominator:
\[ \frac{1}{q + r} + \frac{1}{p + q} = \frac{(p + q) + (q + r)}{(q + r)(p + q)} = \frac{p + 2q + r}{(q + r)(p + q)} \]

4. **Setting the Equation**:
Now substituting back into the A.P. condition:
\[ \frac{2}{r + p} = \frac{p + 2q + r}{(q + r)(p + q)} \]

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

6. **Expanding Both Sides**:
- Left Side:
\[ 2(qp + q^2 + pr + qr) = 2qp + 2q^2 + 2pr + 2qr \]
- Right Side:
\[ (r + p)(p + 2q + r) = rp + 2rq + r^2 + p^2 + 2pq + pr \]
This simplifies to:
\[ p^2 + 2pq + r^2 + 2rq + 2rp \]

7. **Setting the Equation**:
Now we have:
\[ 2qp + 2q^2 + 2pr + 2qr = p^2 + 2pq + r^2 + 2rq + 2rp \]

8. **Rearranging Terms**:
Rearranging gives:
\[ 2q^2 - p^2 - r^2 + 2qp - 2rp - 2rq = 0 \]

9. **Factoring**:
This can be rearranged and factored to show that:
\[ (p - q)^2 + (q - r)^2 + (r - p)^2 = 0 \]
This implies:
\[ p = q = r \]

### Conclusion:
Thus, \( p^2, q^2, r^2 \) are in A.P. if \( p, q, r \) are equal.