In an H.P., `T_(p) = q (p + q), T_(q) = p (p + q)`, then p and q are the roots of

Options

A. `x^(2) - T_(p + q) x + T_(pq) = 0`
B. `x^(2) - T_(pq) x + T_(p + q) = 0`
C. `x^(2) - 2T_(p + q) x + T_(pq) = 0`
D. `x^(2) - T_(pq) x + 2T_(p + q) = 0`

Answer

To solve the problem, we need to find the roots of a quadratic equation given the terms of a Harmonic Progression (H.P.). We are given:

1. \( T_p = q(p + q) \)
2. \( T_q = p(p + q) \)

Where \( T_p \) and \( T_q \) are the \( p \)-th and \( q \)-th terms of the H.P. respectively.

### Step-by-Step Solution:

**Step 1: Write the formulas for \( T_p \) and \( T_q \)**

In an H.P., the \( n \)-th term can be expressed in terms of the first term \( A \) and the common difference \( D \):
\[ T_n = \frac{1}{\frac{1}{A} + (n-1) \cdot \frac{1}{D}} \]

Thus, we can write:
\[ T_p = \frac{1}{\frac{1}{A} + \frac{p-1}{D}} \quad \text{and} \quad T_q = \frac{1}{\frac{1}{A} + \frac{q-1}{D}} \]

**Step 2: Set up the equations**

From the given information:
\[ \frac{1}{\frac{1}{A} + \frac{p-1}{D}} = q(p + q) \quad \text{(1)} \]
\[ \frac{1}{\frac{1}{A} + \frac{q-1}{D}} = p(p + q) \quad \text{(2)} \]

**Step 3: Rearranging the equations**

From equation (1):
\[ \frac{1}{A} + \frac{p-1}{D} = \frac{1}{q(p + q)} \]
This gives us:
\[ A + (p-1) \cdot D = \frac{D}{q(p + q)} \quad \text{(3)} \]

From equation (2):
\[ \frac{1}{A} + \frac{q-1}{D} = \frac{1}{p(p + q)} \]
This gives us:
\[ A + (q-1) \cdot D = \frac{D}{p(p + q)} \quad \text{(4)} \]

**Step 4: Subtract equations (3) and (4)**

Subtracting equation (4) from equation (3):
\[ (p - q) \cdot D = \frac{D}{q(p + q)} - \frac{D}{p(p + q)} \]
Factoring out \( D \):
\[ (p - q) \cdot D = D \left( \frac{1}{q} - \frac{1}{p} \right) \cdot \frac{1}{(p + q)} \]

**Step 5: Solve for \( D \)**

Assuming \( D \neq 0 \):
\[ p - q = \frac{1}{q} - \frac{1}{p} \]
This simplifies to:
\[ (p - q) \cdot pq = p - q \]
Thus:
\[ D = \frac{1}{pq(p + q)} \]

**Step 6: Find \( A \)**

Substituting \( D \) back into either equation (3) or (4) to find \( A \):
Using equation (3):
\[ A + (p - 1) \cdot \frac{1}{pq(p + q)} = \frac{1}{q(p + q)} \]
This leads to:
\[ A = \frac{1}{pq(p + q)} - (p - 1) \cdot \frac{1}{pq(p + q)} \]
Thus:
\[ A = \frac{1}{pq(p + q)} \cdot (1 - (p - 1)) = \frac{1}{pq(p + q)} \cdot (2 - p) \]

**Step 7: Form the quadratic equation**

Using the values of \( A \) and \( D \), we can find \( T_{p+q} \) and \( T_{pq} \):
\[ T_{p+q} = \frac{1}{A + (p + q - 1)D} \]
\[ T_{pq} = \frac{1}{A + (pq - 1)D} \]

The roots of the quadratic equation are given by:
\[ x^2 - (T_{p+q})x + T_{pq} = 0 \]

Thus, \( p \) and \( q \) are the roots of the equation:
\[ x^2 - (p + q)x + pq = 0 \]

### Final Answer:
The roots \( p \) and \( q \) are the roots of the quadratic equation:
\[ x^2 - (p + q)x + pq = 0 \]