In an A.P. let `T_(r)` denote `r ^(th)` term from beginning, `T_(p) = (1)/(q (p +q)), T_(q) =(1)/(p(p+q)),` then :

Options

A. `T_(1)=` common difference
B. `T_(p+q) =(1)/(pq)`
C. `T_(pq) =(1)/(p+q)`
D. `T_(p+q) =(1)/(p ^(2) q ^(2))`

Answer

To solve the problem, we will follow these steps:

### Step 1: Define the terms in the A.P.
In an arithmetic progression (A.P.), the r-th term \( T_r \) can be expressed as:
\[ T_r = a + (r-1)d \]
where \( a \) is the first term and \( d \) is the common difference.

### Step 2: Write the given terms in terms of \( a \) and \( d \)
We are given:
\[ T_p = \frac{1}{q(p + q)} \quad \text{and} \quad T_q = \frac{1}{p(p + q)} \]
Using the formula for the r-th term, we can write:
\[ T_p = a + (p-1)d = \frac{1}{q(p + q)} \quad \text{(1)} \]
\[ T_q = a + (q-1)d = \frac{1}{p(p + q)} \quad \text{(2)} \]

### Step 3: Set up the equations
From equations (1) and (2), we have:
1. \( a + (p-1)d = \frac{1}{q(p + q)} \)
2. \( a + (q-1)d = \frac{1}{p(p + q)} \)

### Step 4: Subtract the equations
Subtract equation (1) from equation (2):
\[ (a + (q-1)d) - (a + (p-1)d) = \frac{1}{p(p + q)} - \frac{1}{q(p + q)} \]
This simplifies to:
\[ (q - p)d = \frac{1}{p(p + q)} - \frac{1}{q(p + q)} \]
Combining the right-hand side:
\[ (q - p)d = \frac{q - p}{pq(p + q)} \]
Assuming \( q \neq p \), we can divide both sides by \( q - p \):
\[ d = \frac{1}{pq(p + q)} \]

### Step 5: Substitute \( d \) back to find \( a \)
Now, substitute \( d \) back into either equation (1) or (2) to find \( a \). We will use equation (1):
\[ a + (p-1)\left(\frac{1}{pq(p + q)}\right) = \frac{1}{q(p + q)} \]
Multiply through by \( pq(p + q) \) to eliminate the denominator:
\[ a \cdot pq(p + q) + (p-1) = \frac{pq}{q} \]
This simplifies to:
\[ a \cdot pq(p + q) + (p-1) = p \quad \Rightarrow \quad a \cdot pq(p + q) = p - (p - 1) = 1 \]
Thus,
\[ a = \frac{1}{pq(p + q)} \]

### Step 6: Verify the options
Now we need to verify the options given in the problem. We will check:
1. \( T_1 \)
2. \( T_{p+1} \)
3. \( T_{pq} \)

#### Finding \( T_1 \):
\[ T_1 = a = \frac{1}{pq(p + q)} \]

#### Finding \( T_{p+1} \):
\[ T_{p+1} = a + pd = \frac{1}{pq(p + q)} + p\left(\frac{1}{pq(p + q)}\right) = \frac{1 + p}{pq(p + q)} \]

#### Finding \( T_{pq} \):
\[ T_{pq} = a + (pq-1)d = \frac{1}{pq(p + q)} + (pq-1)\left(\frac{1}{pq(p + q)}\right) = \frac{1 + (pq-1)}{pq(p + q)} = \frac{pq}{pq(p + q)} = \frac{1}{p + q} \]

### Conclusion
From our calculations, we can conclude that:
- \( T_1 = d \)
- \( T_{p+1} = \frac{1 + p}{pq(p + q)} \)
- \( T_{pq} = \frac{1}{p + q} \)

Thus, the answer is that the first option is satisfied.