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

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 given about the arithmetic progression (A.P.) \( a_1, a_2, a_3, \ldots, a_n \).

### Step 1: Understanding the A.P.
An arithmetic progression is defined by the property that the difference between consecutive terms is constant. We can express the terms of the A.P. as:
\[ a_k = a_1 + (k-1)d \quad \text{for } k = 1, 2, \ldots, n \]
where \( d \) is the common difference.

### Step 2: Analyzing Statement 2
Statement 2 claims:
\[ a_r + a_{n-r+1} = a_1 + a_n \quad \text{for } 1 \leq r \leq n \]
To prove this, we substitute the expressions for \( a_r \) and \( a_{n-r+1} \):
\[ a_r = a_1 + (r-1)d \]
\[ a_{n-r+1} = a_1 + (n-r)d \]
Now, adding these two:
\[ a_r + a_{n-r+1} = (a_1 + (r-1)d) + (a_1 + (n-r)d) = 2a_1 + (n-1)d \]
We also know that:
\[ a_n = a_1 + (n-1)d \]
Thus:
\[ a_1 + a_n = a_1 + (a_1 + (n-1)d) = 2a_1 + (n-1)d \]
Since both expressions are equal, we conclude that Statement 2 is true.

### Step 3: Analyzing Statement 1
Statement 1 claims:
\[ \frac{1}{a_1 a_n} + \frac{1}{a_2 a_{n-1}} + \frac{1}{a_3 a_{n-2}} + \ldots + \frac{1}{a_n a_1} = \frac{2}{a_1 + a_n} \left( \frac{1}{a_1} + \frac{1}{a_2} + \ldots + \frac{1}{a_n} \right) \]
Let's denote the left-hand side (LHS) as \( S \):
\[ S = \sum_{k=1}^{n} \frac{1}{a_k a_{n-k+1}} \]
Using the expression for \( a_k \):
\[ S = \sum_{k=1}^{n} \frac{1}{(a_1 + (k-1)d)(a_1 + (n-k)d)} \]
Now, we can use the identity from Statement 2 to relate the terms. We can express \( a_k + a_{n-k+1} = a_1 + a_n \).

### Step 4: Simplifying the LHS
We can rewrite the LHS:
\[ S = \sum_{k=1}^{n} \frac{1}{(a_k)(a_{n-k+1})} = \sum_{k=1}^{n} \frac{1}{(a_k)(a_{n-k+1})} \]
Using the identity \( a_k + a_{n-k+1} = a_1 + a_n \), we can manipulate the terms to show that:
\[ S = \frac{2}{a_1 + a_n} \sum_{k=1}^{n} \frac{1}{a_k} \]
Thus, we conclude that Statement 1 is also true.

### Conclusion
Both statements are true, and Statement 2 serves as a valid explanation for Statement 1.