In an A.P., `T_(m) = (1)/(n), T_(n) = (1)/(m)`, then a - d is equal to

Options

A. 0
B. 1
C. `(1)/(mn)`
D. `(1)/(m + n)`

Answer

To solve the problem, we need to find the value of \( a - d \) given that \( T_m = \frac{1}{n} \) and \( T_n = \frac{1}{m} \) in an arithmetic progression (A.P.).

### Step-by-Step Solution:

1. **Write the general term of an A.P.**:
The \( k^{th} \) term of an A.P. can be expressed as:
\[ T_k = a + (k - 1)d \]
where \( a \) is the first term and \( d \) is the common difference.

2. **Set up the equations**:
From the problem, we have:
\[ T_m = a + (m - 1)d = \frac{1}{n} \quad \text{(1)} \]
\[ T_n = a + (n - 1)d = \frac{1}{m} \quad \text{(2)} \]

3. **Subtract the two equations**:
We can subtract equation (1) from equation (2):
\[ (a + (n - 1)d) - (a + (m - 1)d) = \frac{1}{m} - \frac{1}{n} \]
Simplifying this gives:
\[ (n - 1)d - (m - 1)d = \frac{1}{m} - \frac{1}{n} \]
\[ (n - m)d = \frac{1}{m} - \frac{1}{n} \]

4. **Simplify the right-hand side**:
The right-hand side can be simplified:
\[ \frac{1}{m} - \frac{1}{n} = \frac{n - m}{mn} \]
Thus, we have:
\[ (n - m)d = \frac{n - m}{mn} \]

5. **Divide both sides by \( n - m \)** (assuming \( n \neq m \)):
\[ d = \frac{1}{mn} \]

6. **Substitute \( d \) back into one of the original equations**:
Let's substitute \( d \) into equation (1):
\[ a + (m - 1)\left(\frac{1}{mn}\right) = \frac{1}{n} \]
Rearranging gives:
\[ a + \frac{m - 1}{mn} = \frac{1}{n} \]
\[ a = \frac{1}{n} - \frac{m - 1}{mn} \]
\[ a = \frac{1}{n} - \frac{m - 1}{mn} = \frac{m - (m - 1)}{mn} = \frac{1}{mn} \]

7. **Find \( a - d \)**:
Now we can find \( a - d \):
\[ a - d = \frac{1}{mn} - \frac{1}{mn} = 0 \]

### Final Answer:
Thus, \( a - d = 0 \).