If the `m^(th)` term of an H.P. is n and the `n^(th)` term is m, show that its `(mn)^(th)` term is 1.

Answer

To solve the problem, we need to show that the \( (mn)^{th} \) term of a Harmonic Progression (H.P.) is 1, given that the \( m^{th} \) term is \( n \) and the \( n^{th} \) term is \( m \).

### Step-by-Step Solution:

1. **Understanding the Terms in H.P.**:
In a Harmonic Progression, the terms can be expressed in terms of an Arithmetic Progression (A.P.). If the \( m^{th} \) term of an H.P. is \( n \), then the \( m^{th} \) term of the corresponding A.P. is \( \frac{1}{n} \). Similarly, if the \( n^{th} \) term of the H.P. is \( m \), then the \( n^{th} \) term of the corresponding A.P. is \( \frac{1}{m} \).

2. **Setting Up the Equations**:
Let \( a \) be the first term and \( d \) be the common difference of the A.P. Then, we can write:
\[ T_m = a + (m-1)d = \frac{1}{n} \quad \text{(1)} \]
\[ T_n = a + (n-1)d = \frac{1}{m} \quad \text{(2)} \]

3. **Subtracting the Equations**:
Now, we will subtract equation (1) from equation (2):
\[ (a + (n-1)d) - (a + (m-1)d) = \frac{1}{m} - \frac{1}{n} \]
This simplifies to:
\[ (n - m)d = \frac{1}{m} - \frac{1}{n} \]
\[ (n - m)d = \frac{n - m}{mn} \]

4. **Solving for \( d \)**:
Assuming \( n \neq m \) (if \( n = m \), the terms would not be distinct), we can divide by \( (n - m) \):
\[ d = \frac{1}{mn} \]

5. **Finding \( a \)**:
Now, substituting \( d \) back into equation (1):
\[ a + (m - 1)\left(\frac{1}{mn}\right) = \frac{1}{n} \]
Rearranging gives:
\[ a = \frac{1}{n} - \frac{m - 1}{mn} \]
\[ a = \frac{m}{mn} - \frac{m - 1}{mn} = \frac{1}{mn} \]

6. **Finding the \( (mn)^{th} \) Term**:
Now, we can find the \( (mn)^{th} \) term:
\[ T_{mn} = a + (mn - 1)d \]
Substituting the values of \( a \) and \( d \):
\[ T_{mn} = \frac{1}{mn} + (mn - 1)\left(\frac{1}{mn}\right) \]
\[ T_{mn} = \frac{1}{mn} + \frac{mn - 1}{mn} = \frac{1 + (mn - 1)}{mn} = \frac{mn}{mn} = 1 \]

### Conclusion:
Thus, we have shown that the \( (mn)^{th} \) term of the H.P. is indeed \( 1 \).