The mth term of an A.P. is `(1)/(n)` and nth term is `(1)/(m).` Its (mn)th term is :

Options

A. mn
B. `(1)/(mn)`
C. 1
D. None of these

Explanation

To solve the problem step by step, we need to find the (mn)th term of the arithmetic progression (A.P.) given that the mth term is \( \frac{1}{n} \) and the nth term is \( \frac{1}{m} \).

### Step 1: Write the formulas for the mth and nth terms of the A.P.
The mth term \( T_m \) of an A.P. can be expressed as:
\[ T_m = a + (m - 1)d \]
where \( a \) is the first term and \( d \) is the common difference.

Similarly, the nth term \( T_n \) is given by:
\[ T_n = a + (n - 1)d \]

### Step 2: Set up the equations based on the given terms.
From the problem, we have:
1. \( T_m = a + (m - 1)d = \frac{1}{n} \) (Equation 1)
2. \( T_n = a + (n - 1)d = \frac{1}{m} \) (Equation 2)

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

### Step 4: Simplify the right side.
To simplify \( \frac{1}{n} - \frac{1}{m} \), we find a common denominator:
\[ \frac{1}{n} - \frac{1}{m} = \frac{m - n}{mn} \]
Thus, we have:
\[ (m - n)d = \frac{m - n}{mn} \]

### Step 5: Solve for \( d \).
Assuming \( m \neq n \), we can divide both sides by \( m - n \):
\[ d = \frac{1}{mn} \]

### Step 6: Substitute \( d \) back into one of the original equations to find \( a \).
Using Equation 1:
\[ a + (m - 1)d = \frac{1}{n} \]
Substituting \( d \):
\[ a + (m - 1) \cdot \frac{1}{mn} = \frac{1}{n} \]
This simplifies to:
\[ a + \frac{m - 1}{mn} = \frac{1}{n} \]
Now, isolate \( a \):
\[ a = \frac{1}{n} - \frac{m - 1}{mn} \]
Finding a common denominator:
\[ a = \frac{m - 1 - (m - 1)}{mn} = \frac{1}{mn} \]

### Step 7: Find the (mn)th term.
Now, we can find the (mn)th term \( T_{mn} \):
\[ T_{mn} = a + (mn - 1)d \]
Substituting \( a \) and \( d \):
\[ T_{mn} = \frac{1}{mn} + (mn - 1) \cdot \frac{1}{mn} \]
This simplifies to:
\[ T_{mn} = \frac{1}{mn} + \frac{mn - 1}{mn} = \frac{1 + mn - 1}{mn} = \frac{mn}{mn} = 1 \]

### Final Answer:
The (mn)th term of the A.P. is:
\[ \boxed{1} \]