If the pth term of an A.P. is q and the qth term isp, then find its rth term.

Explanation

To solve the problem step by step, we will use the properties of an Arithmetic Progression (A.P.).

### Step 1: Understand the given information
We are given that the pth term of an A.P. is q, and the qth term is p. We need to find the rth term of the A.P.

### Step 2: Write the formulas for the pth and qth terms
The formula for the nth term of an A.P. is given by:
\[ T_n = a + (n - 1)d \]
where \(a\) is the first term and \(d\) is the common difference.

From the problem:
- The pth term \(T_p = q\):
\[ T_p = a + (p - 1)d = q \quad \text{(Equation 1)} \]
- The qth term \(T_q = p\):
\[ T_q = a + (q - 1)d = p \quad \text{(Equation 2)} \]

### Step 3: Set up the equations
We have two equations from the terms:
1. \(a + (p - 1)d = q\)
2. \(a + (q - 1)d = p\)

### Step 4: Subtract Equation 1 from Equation 2
To eliminate \(a\), we can subtract Equation 1 from Equation 2:
\[ (a + (q - 1)d) - (a + (p - 1)d) = p - q \]
This simplifies to:
\[ (q - 1)d - (p - 1)d = p - q \]
\[ (q - p)d = p - q \]

### Step 5: Solve for \(d\)
Rearranging gives:
\[ d(q - p) = -(q - p) \]
Assuming \(q \neq p\), we can divide both sides by \(q - p\):
\[ d = -1 \]

### Step 6: Substitute \(d\) back into one of the equations to find \(a\)
Now, we can substitute \(d = -1\) back into Equation 1:
\[ a + (p - 1)(-1) = q \]
This simplifies to:
\[ a - p + 1 = q \]
So,
\[ a = q + p - 1 \]

### Step 7: Find the rth term
Now we can find the rth term \(T_r\):
\[ T_r = a + (r - 1)d \]
Substituting the values of \(a\) and \(d\):
\[ T_r = (q + p - 1) + (r - 1)(-1) \]
This simplifies to:
\[ T_r = q + p - 1 - (r - 1) \]
\[ T_r = q + p - r \]

### Final Answer
Thus, the rth term of the A.P. is:
\[ \boxed{p + q - r} \]