The sum of the first `(2p+1)` terms of an AP is `{(p+1).(2p+1)}.` Which one of the following inferences can be drawn ?

Options

A. The `(p+1)^(th)` term of the AP is `(2p+1)`
B. The `(2p+1)^(th)` term of the AP is `(2p+1)`
C. The `(2p+1)^(th)` term of the AP is `(p+1)`
D. The `(p+1)^(th)` term of the AP is `(p+1)`

Explanation

To solve the given problem, we need to analyze the information provided about the arithmetic progression (AP) and the sum of its terms.

### Step-by-Step Solution:

1. **Understanding the Given Information**:
We know that the sum of the first `(2p + 1)` terms of an AP is given by:
\[ S_{2p+1} = (p + 1)(2p + 1) \]

2. **Using the Formula for the Sum of an AP**:
The formula for the sum of the first `n` terms of an AP is:
\[ S_n = \frac{n}{2} \left(2a + (n - 1)d\right) \]
For our case, substituting `n = 2p + 1`, we have:
\[ S_{2p+1} = \frac{2p + 1}{2} \left(2a + (2p)d\right) \]

3. **Equating the Two Expressions for the Sum**:
Set the two expressions for the sum equal to each other:
\[ (p + 1)(2p + 1) = \frac{2p + 1}{2} \left(2a + 2pd\right) \]

4. **Simplifying the Equation**:
We can multiply both sides by 2 to eliminate the fraction:
\[ 2(p + 1)(2p + 1) = (2p + 1)(2a + 2pd) \]
Since \(2p + 1\) is common on both sides (and assuming \(2p + 1 \neq 0\)), we can divide both sides by \(2p + 1\):
\[ 2(p + 1) = 2a + 2pd \]

5. **Rearranging the Equation**:
Rearranging gives us:
\[ 2a + 2pd = 2(p + 1) \implies 2a + 2pd = 2p + 2 \]
Dividing through by 2:
\[ a + pd = p + 1 \]

6. **Identifying the Term**:
The term \(a + pd\) represents the \((p + 1)\)th term of the AP. Thus, we conclude:
\[ T_{p + 1} = a + pd = p + 1 \]

### Conclusion:
From the above steps, we can conclude that the \((p + 1)\)th term of the AP is equal to \(p + 1\).

### Final Answer:
The correct inference is that the \((p + 1)\)th term of the AP is \(p + 1\).