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

Options

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

Explanation

To find the sum of the \((p+q)^{th}\) and \((p-q)^{th}\) terms of an Arithmetic Progression (AP), we can follow these steps:

### Step-by-Step Solution:

1. **Define the first term and common difference**:
Let the first term of the AP be \(a\) and the common difference be \(d\).

2. **Find the \((p+q)^{th}\) term**:
The formula for the \(n^{th}\) term of an AP is given by:
\[ T_n = a + (n-1)d \]
Therefore, the \((p+q)^{th}\) term is:
\[ T_{p+q} = a + (p+q-1)d \]

3. **Find the \((p-q)^{th}\) term**:
Similarly, the \((p-q)^{th}\) term is:
\[ T_{p-q} = a + (p-q-1)d \]

4. **Sum the two terms**:
Now, we need to find the sum of these 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 get:
\[ T_{p+q} + T_{p-q} = 2a + \left((p+q-1) + (p-q-1)\right)d \]

5. **Combine like terms**:
Now, simplify the expression inside the parentheses:
\[ (p+q-1) + (p-q-1) = p + q - 1 + p - q - 1 = 2p - 2 \]
Therefore, the sum becomes:
\[ T_{p+q} + T_{p-q} = 2a + (2p - 2)d \]

6. **Factor out the common terms**:
We can factor out 2 from the expression:
\[ T_{p+q} + T_{p-q} = 2\left(a + (p-1)d\right) \]

7. **Conclusion**:
The sum of the \((p+q)^{th}\) and \((p-q)^{th}\) terms of the AP is:
\[ 2\left(a + (p-1)d\right) \]