Sequences & Series Question - None

Question

If $a_1, a_2, a_3, \ldots, a_{4001}$ are terms of an AP such that $\frac{1}{a_1 a_2} + \frac{1}{a_2 a_3} + \ldots + \frac{1}{a_{4000} a_{4001}} = 10$ and $a_2 + a_{4000} = 50$, then $|a_1 - a_{4001}|$ is equal to

Accepted Answer

Step 1: Set up the AP structure

Let the common difference be d. Then a_n = a_1 + (n-1)d for each term.

Step 2: Decompose the general term using partial fractions

For consecutive terms a_n and a_(n+1):

1/(a_n·a_(n+1)) = 1/(a_n(a_n + d))

Using partial fractions: 1/(a_n(a_n + d)) = (1/d)[1/a_n - 1/a_(n+1)]

Step 3: Apply telescoping to the sum

∑(n=1 to 4000) 1/(a_n·a_(n+1)) = (1/d)∑(n=1 to 4000)[1/a_n - 1/a_(n+1)]

= (1/d)[1/a_1 - 1/a_4001]

= (1/d)·(a_4001 - a_1)/(a_1·a_4001)

Step 4: Use the given condition

We know: (1/d)·(a_4001 - a_1)/(a_1·a_4001) = 10

Since a_4001 - a_1 = (4001 - 1)d = 4000d:

(1/d)·(4000d)/(a_1·a_4001) = 10

4000/(a_1·a_4001) = 10

Therefore: a_1·a_4001 = 400

Step 5: Use the second condition

Given: a_2 + a_4000 = 50

a_2 = a_1 + d and a_4000 = a_1 + 3999d

(a_1 + d) + (a_1 + 3999d) = 50

2a_1 + 4000d = 50

a_1 + 2000d = 25

Step 6: Find the middle term

Note that a_2001 = a_1 + 2000d = 25

Also, a_1 + a_4001 = a_1 + (a_1 + 4000d) = 2(a_1 + 2000d) = 2(25) = 50

Step 7: Calculate |a_1 - a_4001|

Let a_1 + a_4001 = 50 and a_1·a_4001 = 400

Using (a_1 - a_4001)² = (a_1 + a_4001)² - 4a_1·a_4001

= (50)² - 4(400)

= 2500 - 1600

= 900

Therefore: |a_1 - a_4001| = √900 = 40

∴ Answer: c

Question:

Step-by-Step Solution

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