Let `a_1, a_2, a_3, ...a_(n)` be an AP. then: `1 / (a_1 a_n) + 1 / (a_2 a_(n-1)) + 1 /(a_3a_(n-2))+......+ 1 /(a_(n) a_1) = `

Options

A. 2
B. `a_(1)+a_(n)`
C. `2(a_(1)+a_(n1))`
D. `(n)/(a_(1)a_(n1))`

Explanation

To solve the problem, we need to evaluate the expression:

\[ S = \frac{1}{a_1 a_n} + \frac{1}{a_2 a_{n-1}} + \frac{1}{a_3 a_{n-2}} + \ldots + \frac{1}{a_n a_1} \]

where \( a_1, a_2, a_3, \ldots, a_n \) are terms of an arithmetic progression (AP).

### Step 1: Define the terms of the AP
Let the first term of the AP be \( a_1 = a \) and the common difference be \( d \). Then, the \( n \)-th term can be expressed as:

\[ a_n = a + (n-1)d \]

### Step 2: Write the general term
The \( k \)-th term of the AP can be expressed as:

\[ a_k = a + (k-1)d \]

### Step 3: Substitute the terms into the expression
Now we can substitute the terms into the expression \( S \):

\[ S = \frac{1}{a (a + (n-1)d)} + \frac{1}{(a + d)(a + (n-2)d)} + \frac{1}{(a + 2d)(a + (n-3)d)} + \ldots + \frac{1}{(a + (n-1)d)a} \]

### Step 4: Pair the terms
Notice that each term can be paired:

\[ S = \left( \frac{1}{a_1 a_n} + \frac{1}{a_2 a_{n-1}} + \ldots + \frac{1}{a_n a_1} \right) \]

This can be rewritten as:

\[ S = \sum_{k=1}^{n} \frac{1}{a_k a_{n-k+1}} \]

### Step 5: Simplify the expression
Using the fact that \( a_k = a + (k-1)d \) and \( a_{n-k+1} = a + (n-k)d \):

\[ S = \sum_{k=1}^{n} \frac{1}{(a + (k-1)d)(a + (n-k)d)} \]

### Step 6: Factor out common terms
Notice that:

\[ a + (n-k)d = a + (n-1)d - (k-1)d \]

Thus, we can express \( S \) in terms of \( a \) and \( d \).

### Step 7: Use the sum of reciprocals
The sum of the reciprocals of the terms in an AP can be simplified. The total number of terms is \( n \), and the average of the terms is given by:

\[ \text{Average} = \frac{a_1 + a_n}{2} = \frac{2a + (n-1)d}{2} \]

### Final Result
The final result can be expressed as:

\[ S = \frac{n}{\frac{2a + (n-1)d}{2}} = \frac{2n}{2a + (n-1)d} \]

### Conclusion
Thus, the value of the expression is:

\[ S = \frac{n}{a + \frac{(n-1)d}{2}} \]