The sum of `(p+q)^(th) and (p-q)^(th)` terms of an AP equal to

Options

A. A)`(2p)^(th)` term
B. B)`(2q)^(th)` term
C. C)Twice the `p^(th)` term
D. D)Twice the `q^(th)` term

Answer

To solve the problem, we need to find the sum of the \( (p+q)^{th} \) and \( (p-q)^{th} \) terms of an arithmetic progression (AP). Let's denote the first term of the AP as \( A \) and the common difference as \( D \).

### Step-by-Step Solution:

1. **Identify the Terms**:
- The \( n^{th} \) term of an AP can be expressed as:
\[ T_n = A + (n-1)D \]
- Therefore, the \( (p+q)^{th} \) term is:
\[ T_{p+q} = A + (p+q-1)D \]
- The \( (p-q)^{th} \) term is:
\[ T_{p-q} = A + (p-q-1)D \]

2. **Sum the Terms**:
- Now, let's add the two terms:
\[ T_{p+q} + T_{p-q} = \left( A + (p+q-1)D \right) + \left( A + (p-q-1)D \right) \]
- Simplifying this, we have:
\[ = A + (p+q-1)D + A + (p-q-1)D \]
- Combining like terms:
\[ = 2A + \left( (p+q-1) + (p-q-1) \right)D \]
- Simplifying the expression inside the parentheses:
\[ = 2A + \left( p + q - 1 + p - q - 1 \right)D \]
\[ = 2A + (2p - 2)D \]
\[ = 2A + 2(p-1)D \]

3. **Factor Out the 2**:
- We can factor out the 2:
\[ = 2 \left( A + (p-1)D \right) \]
- Recognizing that \( A + (p-1)D \) is the \( p^{th} \) term of the AP:
\[ = 2T_p \]

### Conclusion:
The sum of the \( (p+q)^{th} \) and \( (p-q)^{th} \) terms of an AP is equal to:
\[ 2T_p \]
Thus, the correct answer is that the sum equals twice the \( p^{th} \) term of the AP.