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}} + ... + \frac{1}{a_{4000}a_{4001}} = 10\$ and $a_{2} + a_{4000} = 50,$ then $|a_{1} - a_{4001}|$ is equal to

Question

Dr. Abubacker · Accepted Answer

(b)
Now, $\frac{1}{a_{1}a_{2}} + \frac{1}{a_{2}a_{3}} + ... + \frac{1}{a_{4000}a_{4001}}$
$= \frac{1}{d}\left( \frac{a_{2} - a_{1}}{a_{1}a_{2}} + \frac{a_{3} - a_{2}}{a_{2}a_{3}} + ... + \frac{a_{4001} - a_{4000}}{a_{4000}a_{4001}} ight)$
$= \frac{1}{d}\left( \frac{1}{a_{1}} - \frac{1}{a_{2}} + \frac{1}{a_{2}} - \frac{1}{a_{3}} + ... + \frac{1}{a_{4000}} - \frac{1}{a_{4001}} ight)$
$= \frac{1}{d}\left( \frac{1}{a_{1}} - \frac{1}{a_{4001}} ight) = \frac{400}{a_{1}a_{4001}} = 10$ (given)
$\Rightarrow \ \ a_{1}a_{4001} = 400\$ …(i)
$a_{1} + a_{4001} = a_{2} + a_{4000} = 50$ …(ii)
$	herefore\ \ \left( a_{1} - a_{4001} ight)^{2} = \left( a_{1} + a_{4000} ight)^{2} - 4a_{1}a_{4001}$
$= (50)^{2} - 1600$
$\Rightarrow \left| a_{1} - a_{4001} ight| = 30$

Q. 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}} + ... + \frac{1}{a_{4000}a_{4001}} = 10\$ and $a_{2} + a_{4000} = 50,$ then $|a_{1} - a_{4001}|$ is equal to

Explanation

Reviewed by Dr. Abubacker

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