Q. If $a_{1},\ a_{2},\ldots,\ a_{n + 1}$ are in AP, then $\frac{1}{a_{1}a_{2}} + \frac{1}{a_{2}a_{3}} + ... + \frac{1}{a_{n}a_{n + 1}}$ is
A) $\frac{n - 1}{a_{1}a_{n + 1}}$
B) $\frac{1}{a_{1}a_{n + 1}}$
C) $\frac{n + 1}{a_{1}a_{n + 1}}$
D) $\frac{n}{a_{1}a_{n + 1}}$
Explanation
(d)
$\because a_{1},\ a_{2},\ldots,a_{n + 1}$ are in AP and common difference =$d$
Let $S = \frac{1}{a_{1}a_{2}} + \frac{1}{a_{2}a_{3}} + ... + \frac{1}{a_{n}a_{n + 1}}$
$= \frac{1}{d}\ \left\{ \frac{a_{2} - a_{1}}{a_{1}a_{2}} + \frac{a_{3} - a_{2}}{a_{2}a_{3}} + ... + \frac{a_{n + 1} - a_{n}}{a_{n}a_{n + 1}} \right\}$
$= \frac{1}{d}\left\{ \frac{1}{a_{1}} - \frac{1}{a_{2}} + \frac{1}{a_{2}} - \frac{1}{a_{3}} + ... + \frac{1}{a_{n}} - \frac{1}{a_{n + 1}} \right\}$
$= \frac{1}{d}\left\{ \frac{a_{n + 1} - a_{1}}{a_{1}a_{n + 1}} \right\}$
$= \ \frac{nd}{d(a_{1}a_{n + 1})} = \frac{n}{a_{1}a_{n + 1}}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.