The `m^(th)` term of an A.P. is n and its `n^(th)` term is m. Its `p^(th)` term is

Options

A. m + n + p
B. m + n - p
C. m - n + p
D. none

Answer

To solve the problem, we need to find the `p^(th)` term of an arithmetic progression (A.P.) given that the `m^(th)` term is `n` and the `n^(th)` term is `m`.

### Step-by-step Solution:

1. **Define the first term and common difference**:
Let the first term of the A.P. be `a` and the common difference be `d`.

2. **Write the expressions for the `m^(th)` and `n^(th)` terms**:
- The `m^(th)` term can be expressed as:
\[ a_m = a + (m - 1)d = n \quad \text{(Equation 1)} \]
- The `n^(th)` term can be expressed as:
\[ a_n = a + (n - 1)d = m \quad \text{(Equation 2)} \]

3. **Subtract Equation 1 from Equation 2**:
\[ (a + (n - 1)d) - (a + (m - 1)d) = m - n \]
Simplifying this gives:
\[ (n - 1)d - (m - 1)d = m - n \]
\[ (n - m)d = m - n \]
Rearranging gives:
\[ d = \frac{m - n}{n - m} = -1 \]

4. **Substitute the value of `d` back into either equation to find `a`**:
Using Equation 1:
\[ a + (m - 1)(-1) = n \]
Simplifying:
\[ a - m + 1 = n \]
\[ a = n + m - 1 \]

5. **Find the `p^(th)` term**:
The `p^(th)` term can be expressed as:
\[ a_p = a + (p - 1)d \]
Substituting the values of `a` and `d`:
\[ a_p = (n + m - 1) + (p - 1)(-1) \]
Simplifying:
\[ a_p = n + m - 1 - (p - 1) \]
\[ a_p = n + m - p \]

Thus, the `p^(th)` term of the A.P. is:
\[ \boxed{m + n - p} \]