If $a_{1},a_{2},a_{3},\ldots,a_{n}$ be an AP of non-zero terms, then $\frac{1}{a_{1}a_{2}} + \frac{1}{a_{2}a_{3}} + ... + \frac{1}{a_{n - 1}a_{n}}$ is equal to

Question

Dr. Abubacker · Accepted Answer

(a)
Let $a_{2} - a_{1} = a_{3} - a_{2} = ... = a_{n} - a_{n - 1} = d$
$	herefore\ \frac{1}{d}\left( \frac{d}{a_{1}a_{2}} + \frac{d}{a_{2}a_{3}} + ... + \frac{d}{a_{n - 1}a_{n}} ight)$
$= \frac{1}{d}\left\lbrack \frac{a_{2} - a_{1}}{a_{1}a_{2}} + \frac{a_{3} - a_{2}}{a_{2}a_{3}} + ... + \frac{a_{n} - a_{n - 1}}{a_{n - 1}a_{n}} ightbrack$
$= \frac{1}{d}\left\lbrack \frac{1}{a_{1}} - \frac{1}{a_{2}} + \frac{1}{a_{2}} - \frac{1}{a_{3}} + ... + \frac{1}{a_{n - 1}} - \frac{1}{a_{n}} ightbrack$
$= \frac{1}{d}\left\lbrack \frac{1}{a_{1}} - \frac{1}{a_{n}} ightbrack = \frac{1}{d}\left\lbrack \frac{a_{n} - a_{1}}{a_{1}a_{n}} ightbrack$
$= \frac{n - 1}{a_{1}a_{n}}$

Q. If $a_{1},a_{2},a_{3},\ldots,a_{n}$ be an AP of non-zero terms, then $\frac{1}{a_{1}a_{2}} + \frac{1}{a_{2}a_{3}} + ... + \frac{1}{a_{n - 1}a_{n}}$ is equal to

Explanation

Reviewed by Dr. Abubacker

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