If the `m^(th)` term of an H.P. is n and `n^(th)` term be m, then `(m + n)^(th)` term is

Options

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

Answer

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

### Step-by-Step Solution:

1. **Understanding H.P.**:
The \(m^{th}\) term of an H.P. can be expressed in terms of its corresponding Arithmetic Progression (A.P.). If the A.P. is given by \(a, a+d, a+2d, \ldots\), then the H.P. can be expressed as:
\[ \frac{1}{a}, \frac{1}{a+d}, \frac{1}{a+2d}, \ldots \]

2. **Expressing the \(m^{th}\) term**:
The \(m^{th}\) term of the H.P. is:
\[ T_m = \frac{1}{a + (m-1)d} = n \]
Rearranging gives us:
\[ a + (m-1)d = \frac{1}{n} \quad \text{(Equation 1)} \]

3. **Expressing the \(n^{th}\) term**:
Similarly, the \(n^{th}\) term of the H.P. is:
\[ T_n = \frac{1}{a + (n-1)d} = m \]
Rearranging gives us:
\[ a + (n-1)d = \frac{1}{m} \quad \text{(Equation 2)} \]

4. **Subtracting the two 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:
\[ (n - m)d = \frac{1}{m} - \frac{1}{n} \]
Rearranging gives:
\[ (n - m)d = \frac{n - m}{mn} \]
Assuming \(n \neq m\), we can divide both sides by \(n - m\):
\[ d = \frac{1}{mn} \]

5. **Finding \(a\)**:
Substitute \(d\) back into Equation 1 to find \(a\):
\[ a + (m-1)\frac{1}{mn} = \frac{1}{n} \]
Rearranging gives:
\[ a = \frac{1}{n} - \frac{m-1}{mn} = \frac{m - 1}{mn} \quad \text{(Equation 3)} \]

6. **Finding the \((m+n)^{th}\) term**:
Now we can find the \((m+n)^{th}\) term:
\[ T_{m+n} = \frac{1}{a + (m+n-1)d} \]
Substitute \(a\) and \(d\):
\[ T_{m+n} = \frac{1}{\frac{m-1}{mn} + (m+n-1)\frac{1}{mn}} \]
Simplifying gives:
\[ T_{m+n} = \frac{1}{\frac{m-1 + (m+n-1)}{mn}} = \frac{mn}{m+n} \]

### Final Answer:
Thus, the \((m+n)^{th}\) term of the H.P. is:
\[ \frac{mn}{m+n} \]