Let `a_1,a_2,a_3` …. `a_n` be in A.P. If `1/(a_1a_n)+1/(a_2a_(n-1))`+… + `1/(a_n a_1) = k/(a_1 + a_n) (1/a_1 + 1/a_2 + …. 1/a_n)` , then k is equal to :

Options

A. 1
B. 2
C. 3
D. 4

Answer

To solve the problem, we need to analyze the given equation involving the terms of an arithmetic progression (A.P.) and simplify it step by step.

### Step 1: Understanding the terms in A.P.
Let the first term of the A.P. be \( a_1 = a \) and the common difference be \( d \). Then the terms of the A.P. can be expressed as:
- \( a_1 = a \)
- \( a_2 = a + d \)
- \( a_3 = a + 2d \)
- ...
- \( a_n = a + (n-1)d \)

### Step 2: Rewrite the left-hand side
We need to evaluate the left-hand side of the equation:
\[ \frac{1}{a_1 a_n} + \frac{1}{a_2 a_{n-1}} + \ldots + \frac{1}{a_n a_1} \]
This can be rewritten using the expressions for \( a_i \):
\[ \frac{1}{a(a + (n-1)d)} + \frac{1}{(a + d)(a + (n-2)d)} + \ldots + \frac{1}{(a + (n-1)d)a} \]

### Step 3: Generalize the terms
Notice that each term in the series can be expressed as:
\[ \frac{1}{a_i a_{n-i+1}} = \frac{1}{(a + (i-1)d)(a + (n-i)d)} \]
for \( i = 1, 2, \ldots, n \).

### Step 4: Factor out common terms
We can factor out the common term \( a + a_n \) from the denominator:
\[ = \frac{1}{(a + a_n)} \left( \frac{1}{a} + \frac{1}{a + d} + \ldots + \frac{1}{a + (n-1)d} \right) \]
This gives us:
\[ \frac{1}{(a + a_n)} \sum_{i=1}^{n} \frac{1}{a + (i-1)d} \]

### Step 5: Simplifying the right-hand side
The right-hand side of the equation is:
\[ \frac{k}{a_1 + a_n} \left( \frac{1}{a_1} + \frac{1}{a_2} + \ldots + \frac{1}{a_n} \right) \]
Substituting \( a_1 = a \) and \( a_n = a + (n-1)d \), we have:
\[ = \frac{k}{(a + (a + (n-1)d))} \left( \frac{1}{a} + \frac{1}{a + d} + \ldots + \frac{1}{a + (n-1)d} \right) \]

### Step 6: Equating both sides
Now we equate the left-hand side and right-hand side:
\[ \frac{1}{(a + a_n)} \sum_{i=1}^{n} \frac{1}{a + (i-1)d} = \frac{k}{(a + a_n)} \left( \sum_{i=1}^{n} \frac{1}{a + (i-1)d} \right) \]

### Step 7: Solving for \( k \)
Since both sides have the common term \( \sum_{i=1}^{n} \frac{1}{a + (i-1)d} \) and \( a + a_n \) (which is the same), we can cancel them out, leading to:
\[ 1 = k \]

### Conclusion
Thus, the value of \( k \) is:
\[ \boxed{2} \]