Let `a_(1)+a_(2)+a_(3), . . . ,a_(n-1),a_(n)` be an A.P.
Statement -1: `a_(1)+a_(2)+a_(3)+ . . . +a_(n)=(n)/(2)(a_(1)+a_(n))`
Statement -2 `a_(k)+a_(n-k+1)=a_(1)+a_(n)" for "k=1,2,3, . . . ,` n

Options

A. Statement -1 is true, Statement -2 is True, Statement -2 is a correct explanation for Statement for Statement -1.
B. Statement -1 is true, Statement -2 is True, Statement -2 is not a correct explanation for Statement for Statement -1.
C. Statement -1 is true, Statement -2 is False.
D. Statement -1 is False, Statement -2 is True.

Explanation

To solve the problem, we need to analyze the two statements regarding the arithmetic progression (A.P.) defined by the terms \( a_1, a_2, a_3, \ldots, a_n \).

### Step 1: Verify Statement 1
**Statement 1**: The sum of the first \( n \) terms of an A.P. is given by the formula:
\[ S_n = \frac{n}{2} (a_1 + a_n) \]

**Proof**:
1. The sum of the first \( n \) terms of an A.P. can be expressed as:
\[ S_n = a_1 + a_2 + a_3 + \ldots + a_n \]
2. If we write the sum in reverse order, we have:
\[ S_n = a_n + a_{n-1} + a_{n-2} + \ldots + a_1 \]
3. Adding these two equations gives:
\[ 2S_n = (a_1 + a_n) + (a_2 + a_{n-1}) + (a_3 + a_{n-2}) + \ldots \]
4. Each pair \( (a_k + a_{n-k+1}) \) sums to \( a_1 + a_n \) and there are \( n \) such pairs.
5. Therefore, we can express this as:
\[ 2S_n = n(a_1 + a_n) \]
6. Dividing both sides by 2 gives:
\[ S_n = \frac{n}{2}(a_1 + a_n) \]

Thus, **Statement 1 is true**.

### Step 2: Verify Statement 2
**Statement 2**: For \( k = 1, 2, 3, \ldots, n \), it states:
\[ a_k + a_{n-k+1} = a_1 + a_n \]

**Proof**:
1. We know that in an A.P., the \( k \)-th term can be expressed as:
\[ a_k = a_1 + (k-1)d \]
where \( d \) is the common difference.
2. Similarly, the \( (n-k+1) \)-th term is:
\[ a_{n-k+1} = a_1 + (n-k)d \]
3. Adding these two terms gives:
\[ a_k + a_{n-k+1} = [a_1 + (k-1)d] + [a_1 + (n-k)d] \]
Simplifying this, we get:
\[ a_k + a_{n-k+1} = 2a_1 + (n-1)d \]
4. From the formula for the \( n \)-th term, we have:
\[ a_n = a_1 + (n-1)d \]
5. Therefore, we can substitute \( a_n \) into our equation:
\[ a_k + a_{n-k+1} = 2a_1 + (n-1)d = a_1 + a_n \]

Thus, **Statement 2 is also true**.

### Conclusion
Both statements are true, and Statement 2 correctly explains Statement 1.