In an AP the pth term is q and the `(p+q)`th term is zero, then the qth term is (i) `-p` (ii) `p` (iii) `p+q` (iv) `p-q`

Options

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

Answer

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

### Step 1: Write the formula for the pth term
The pth term of an AP can be expressed as:
\[ a_p = a + (p - 1)d \]
where \(a\) is the first term and \(d\) is the common difference. According to the problem, the pth term is given as \(q\):
\[ a + (p - 1)d = q \quad \text{(Equation 1)} \]

### Step 2: Write the formula for the (p + q)th term
Similarly, the (p + q)th term can be expressed as:
\[ a_{p+q} = a + (p + q - 1)d \]
According to the problem, this term is equal to zero:
\[ a + (p + q - 1)d = 0 \quad \text{(Equation 2)} \]

### Step 3: Set up the equations
Now we have two equations:
1. \(a + (p - 1)d = q\)
2. \(a + (p + q - 1)d = 0\)

### Step 4: Subtract Equation 1 from Equation 2
Subtract Equation 1 from Equation 2:
\[ (a + (p + q - 1)d) - (a + (p - 1)d) = 0 - q \]
This simplifies to:
\[ (p + q - 1)d - (p - 1)d = -q \]
\[ (q)d = -q \]

### Step 5: Solve for d
From the equation \(qd = -q\), we can divide both sides by \(q\) (assuming \(q \neq 0\)):
\[ d = -1 \]

### Step 6: Substitute d back into Equation 1
Now substitute \(d = -1\) back into Equation 1:
\[ a + (p - 1)(-1) = q \]
This simplifies to:
\[ a - (p - 1) = q \]
\[ a = q + p - 1 \quad \text{(Equation 3)} \]

### Step 7: Find the qth term
Now, we need to find the qth term \(a_q\):
\[ a_q = a + (q - 1)d \]
Substituting \(d = -1\) and using Equation 3:
\[ a_q = (q + p - 1) + (q - 1)(-1) \]
This simplifies to:
\[ a_q = (q + p - 1) - (q - 1) \]
\[ = q + p - 1 - q + 1 \]
\[ = p \]

### Conclusion
Thus, the qth term is:
\[ \boxed{p} \]