Arithmetic Progression problem

[utkarshakash]

Homework Statement

Let a₁,a₂,a₃...,a₄₀₀₁ are in A.P. such that $\dfrac{1}{a_1a_2}+\dfrac{1}{a_2a_3}+.......\dfrac{1}{a_{4000}a_{4001}} = 10$ and a₂+a₄₀₀₀=50. Then |a₁-a₄₀₀₁|

The Attempt at a Solution

$\dfrac{1}{a_2} \left( \dfrac{1}{a_1} + \dfrac{1}{a_3} \right) + \dfrac{1}{a_4} \left( \dfrac{1}{a_3} + \dfrac{1}{a_5} \right) ... \\<br /> <br /> 2 \left( \dfrac{1}{a_1a_3} + \dfrac{1}{a_3a_5} ... \right)$

Similarly the number of terms will keep on reducing but it's really difficult to see manually what will I end up with?

 

[Saitama]

Homework Statement

Let a₁,a₂,a₃...,a₄₀₀₁ are in A.P. such that $\dfrac{1}{a_1a_2}+\dfrac{1}{a_2a_3}+.......\dfrac{1}{a_{4000}a_{4001}} = 10$ and a₂+a₄₀₀₀=50. Then |a₁-a₄₀₀₁|

The Attempt at a Solution

$\dfrac{1}{a_2} \left( \dfrac{1}{a_1} + \dfrac{1}{a_3} \right) + \dfrac{1}{a_4} \left( \dfrac{1}{a_3} + \dfrac{1}{a_5} \right) ... \\<br /> <br /> 2 \left( \dfrac{1}{a_1a_3} + \dfrac{1}{a_3a_5} ... \right)$

Similarly the number of terms will keep on reducing but it's really difficult to see manually what will I end up with?

Hi utkarshakash!

Try the following, I am not sure if this is going to work:
$$\frac{1}{a_n a_{n+1}}=\frac{1}{d}\left(\frac{a_{n+1}-a_n}{a_n a_{n+1}}\right)=\frac{1}{d}\left(\frac{1}{a_n}-\frac{1}{a_{n+1}}\right)$$
where $d$ is the common difference of AP.

 

[utkarshakash]

Hi utkarshakash!

Try the following, I am not sure if this is going to work:
$$\frac{1}{a_n a_{n+1}}=\frac{1}{d}\left(\frac{a_{n+1}-a_n}{a_n a_{n+1}}\right)=\frac{1}{d}\left(\frac{1}{a_n}-\frac{1}{a_{n+1}}\right)$$
where $d$ is the common difference of AP.

It does help. Thanks a lot!