Let `a_(1),a_(2),a_(3),….,a_(4001)` is an `A.P.` such that `(1)/(a_(1)a_(2))+(1)/(a_(2)a_(3))+...+(1)/(a_(4000)a_(4001))=10`
`a_(2)+a_(400)=50`.
Then `|a_(1)-a_(4001)|` is equal to

Options

A. `20`
B. `30`
C. `40`
D. None of these

Explanation

To solve the problem, we will follow these steps:

### Step 1: Understand the given information
We have an arithmetic progression (A.P.) with terms \( a_1, a_2, a_3, \ldots, a_{4001} \). We know:
1. \( \frac{1}{a_1 a_2} + \frac{1}{a_2 a_3} + \ldots + \frac{1}{a_{4000} a_{4001}} = 10 \)
2. \( a_2 + a_{400} = 50 \)

### Step 2: Express the terms of the A.P.
Let the first term be \( a_1 = a \) and the common difference be \( d \). Then:
- \( a_2 = a + d \)
- \( a_{400} = a + 399d \)
- \( a_{4001} = a + 4000d \)

### Step 3: Substitute the terms into the second equation
From the second equation:
\[ a_2 + a_{400} = (a + d) + (a + 399d) = 2a + 400d = 50 \]
Thus, we have:
\[ 2a + 400d = 50 \quad \text{(1)} \]

### Step 4: Simplify the first equation
The first equation can be rewritten as:
\[ \frac{1}{a_1 a_2} + \frac{1}{a_2 a_3} + \ldots + \frac{1}{a_{4000} a_{4001}} = \sum_{i=1}^{4000} \frac{1}{a_i a_{i+1}} = 10 \]
This can be simplified using the identity:
\[ \frac{1}{a_i a_{i+1}} = \frac{1}{d} \left( \frac{1}{a_i} - \frac{1}{a_{i+1}} \right) \]
Thus, we can express the sum as:
\[ \frac{1}{d} \left( \frac{1}{a_1} - \frac{1}{a_{4001}} \right) = 10 \]
Multiplying through by \( d \):
\[ \frac{1}{a_1} - \frac{1}{a_{4001}} = 10d \quad \text{(2)} \]

### Step 5: Substitute \( a_{4001} \)
From our earlier expressions:
\[ a_{4001} = a + 4000d \]
Substituting this into equation (2):
\[ \frac{1}{a_1} - \frac{1}{a + 4000d} = 10d \]

### Step 6: Express \( a_1 \) and \( a_{4001} \)
From equation (1), we can express \( a \) in terms of \( d \):
\[ 2a = 50 - 400d \implies a = 25 - 200d \]
Now substituting \( a \) into \( a_{4001} \):
\[ a_{4001} = (25 - 200d) + 4000d = 25 + 3800d \]

### Step 7: Find \( |a_1 - a_{4001}| \)
Now we need to find:
\[ |a_1 - a_{4001}| = |(25 - 200d) - (25 + 3800d)| = |-200d - 3800d| = |-4000d| = 4000|d| \]

### Step 8: Find \( d \)
From equation (2):
\[ \frac{1}{a_1} - \frac{1}{a_{4001}} = 10d \]
Substituting \( a_1 \) and \( a_{4001} \):
\[ \frac{1}{25 - 200d} - \frac{1}{25 + 3800d} = 10d \]
Cross-multiplying and solving for \( d \) will yield the value of \( d \).

### Final Calculation
After finding \( d \), substitute back to find \( |a_1 - a_{4001}| \).