Let `T` be the `r`th term of an A.P. whose first term is `a` and conmon difference is `d`. If for some positive integers `m ,n,` `T_(n)= (1)/(m) , T_(m) = (1)/(n) ` then `(a – d)` equals

Answer

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

### Step-by-Step Solution:

1. **Write the formula for the r-th term of an A.P.:**
The r-th term \( T_r \) of an A.P. is given by:
\[ T_r = a + (r - 1)d \]

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

3. **Subtract Equation 2 from Equation 1:**
\[ (a + (n - 1)d) - (a + (m - 1)d) = \frac{1}{m} - \frac{1}{n} \]
This simplifies to:
\[ (n - 1)d - (m - 1)d = \frac{1}{m} - \frac{1}{n} \]
\[ (n - m)d = \frac{1}{m} - \frac{1}{n} \]

4. **Simplify the right-hand side:**
The right-hand side can be simplified as:
\[ \frac{1}{m} - \frac{1}{n} = \frac{n - m}{mn} \]
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 Equation 1 to find \( a \):**
Substitute \( d = \frac{1}{mn} \) into Equation 1:
\[ a + (n - 1)\left(\frac{1}{mn}\right) = \frac{1}{m} \]
This can be rearranged to find \( a \):
\[ a + \frac{n - 1}{mn} = \frac{1}{m} \]
\[ a = \frac{1}{m} - \frac{n - 1}{mn} \]
\[ a = \frac{n - 1 - n + 1}{mn} = \frac{1}{mn} \]

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

### Final Answer:
Thus, the value of \( a - d \) is:
\[ \boxed{0} \]