Let T, be the rth term of an A.P. for r = 1, 2, 3 … if the some positive integers m,n we have `T_m = 1/n and T_n = 1/m ` then `T_(mn)` equals :

Options

A. a)`1/(mn)`
B. b)`1/m + 1/n`
C. c)1
D. d)0

Answer

To solve the problem, we need to find the value of \( T_{mn} \) given the conditions \( T_m = \frac{1}{n} \) and \( T_n = \frac{1}{m} \), where \( T_r \) is the r-th term of an arithmetic progression (A.P.).

### Step-by-step Solution:

1. **Understanding the A.P. Terms**:
The r-th term of an A.P. can be expressed as:
\[ T_r = a + (r - 1)d \]
where \( a \) is the first term and \( d \) is the common difference.

2. **Setting up the equations**:
From the given conditions:
- For \( T_m \):
\[ T_m = a + (m - 1)d = \frac{1}{n} \quad \text{(1)} \]
- For \( T_n \):
\[ T_n = a + (n - 1)d = \frac{1}{m} \quad \text{(2)} \]

3. **Subtracting the equations**:
We can subtract equation (1) from equation (2):
\[ (a + (n - 1)d) - (a + (m - 1)d) = \frac{1}{m} - \frac{1}{n} \]
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. **Finding the common difference \( d \)**:
The right-hand side can be simplified:
\[ \frac{1}{m} - \frac{1}{n} = \frac{n - m}{mn} \]
Thus, we have:
\[ (n - m)d = \frac{n - m}{mn} \]
Assuming \( n \neq m \), we can divide both sides by \( n - m \):
\[ d = \frac{1}{mn} \]

5. **Finding the first term \( a \)**:
Now, substitute \( d \) back into one of the original equations, let's use equation (1):
\[ a + (m - 1)\left(\frac{1}{mn}\right) = \frac{1}{n} \]
Rearranging gives:
\[ a + \frac{m - 1}{mn} = \frac{1}{n} \]
Multiplying through by \( mn \) to eliminate the fraction:
\[ amn + (m - 1) = m \]
Therefore:
\[ amn = m - (m - 1) = 1 \implies a = \frac{1}{mn} \]

6. **Finding \( T_{mn} \)**:
Now we can find \( T_{mn} \):
\[ T_{mn} = a + (mn - 1)d \]
Substituting \( a \) and \( d \):
\[ T_{mn} = \frac{1}{mn} + (mn - 1)\left(\frac{1}{mn}\right) \]
Simplifying this:
\[ T_{mn} = \frac{1}{mn} + \frac{mn - 1}{mn} = \frac{1 + mn - 1}{mn} = \frac{mn}{mn} = 1 \]

### Final Answer:
Thus, the value of \( T_{mn} \) is:
\[ \boxed{1} \]