If the pth term of an A.P. is q and qth term is p then its rth term will be

Options

A. `p-q+r`
B. `p+q-r`
C. `p+r-q`
D. `p-q-r`

Answer

To solve the problem, we need to find the rth term of an arithmetic progression (A.P.) given that the pth term is q and the qth term is p.

### Step-by-Step Solution:

1. **Understanding the nth term of an A.P.**:
The nth term of an A.P. can be expressed as:
\[ A_n = A + (n - 1) \cdot d \]
where \( A \) is the first term, \( d \) is the common difference, and \( n \) is the term number.

2. **Setting up the equations**:
- From the problem, the pth term is q:
\[ A + (p - 1) \cdot d = q \quad \text{(Equation 1)} \]
- The qth term is p:
\[ A + (q - 1) \cdot d = p \quad \text{(Equation 2)} \]

3. **Subtracting the equations**:
We can subtract Equation 2 from Equation 1:
\[ [A + (p - 1) \cdot d] - [A + (q - 1) \cdot d] = q - p \]
Simplifying this gives:
\[ (p - 1) \cdot d - (q - 1) \cdot d = q - p \]
\[ d \cdot [(p - 1) - (q - 1)] = q - p \]
\[ d \cdot (p - q) = q - p \]

4. **Solving for d**:
Rearranging the equation gives:
\[ d = \frac{q - p}{p - q} = -1 \]

5. **Finding A**:
Now we substitute \( d = -1 \) back into Equation 1 to find \( A \):
\[ A + (p - 1)(-1) = q \]
\[ A - (p - 1) = q \]
\[ A = q + (p - 1) = p + q - 1 \]

6. **Finding the rth term**:
Now we can find the rth term using the formula for the nth term:
\[ A_r = A + (r - 1) \cdot d \]
Substituting \( A \) and \( d \):
\[ A_r = (p + q - 1) + (r - 1)(-1) \]
\[ A_r = (p + q - 1) - (r - 1) \]
\[ A_r = p + q - r \]

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