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If ${a_1},\,{a_2},....,{a_{n + 1}}$ are in $A.P.$, then $\frac{1}{{{a_1}{a_2}}} + \frac{1}{{{a_2}{a_3}}} + ..... + \frac{1}{{{a_n}{a_{n + 1}}}}$ is
$\frac{{n - 1}}{{{a_1}{a_{n + 1}}}}$
$\frac{1}{{{a_1}{a_{n + 1}}}}$
$\frac{{n + 1}}{{{a_1}{a_{n + 1}}}}$
$\frac{n}{{{a_1}{a_{n + 1}}}}$
Solution
(d) ${a_1},{a_2},{a_3},…….,{a_{n + 1}}$ are in $A.P.$
and common difference $= d$
Let $S = \frac{1}{{{a_1}{a_2}}} + \frac{1}{{{a_2}{a_3}}} + ………. + \frac{1}{{{a_n}{a_{n + 1}}}}$
$⇒$ $S = \frac{1}{d}\,\left\{ {\frac{d}{{{a_1}{a_2}}} + \frac{d}{{{a_2}{a_3}}} + …… + \frac{d}{{{a_n}\,\,{a_{n + 1}}}}} \right\}$
$⇒$ $S = \frac{1}{d}\,\left\{ {\frac{{{a_2} – {a_1}}}{{{a_1}{a_2}}} + \frac{{{a_3} – {a_2}}}{{{a_2}{a_3}}} + …… + \frac{{{a_{n + 1}} – {a_n}}}{{{a_n}\,\,\,{a_{n + 1}}}}} \right\}$
$⇒$ $S = \frac{1}{d}\left\{ {\frac{1}{{{a_1}}} – \frac{1}{{{a_2}}} + \frac{1}{{{a_2}}} – \frac{1}{{{a_3}}} + ……. + \frac{1}{{{a_n}}} – \frac{1}{{{a_{n + 1}}}}} \right\}$
$⇒$ $S = \frac{1}{d}\left\{ {\frac{1}{{{a_n}}} – \frac{1}{{{a_{n + 1}}}}} \right\} = \frac{1}{d}\left\{ {\frac{{{a_{n + 1}} – {a_1}}}{{{a_1}{a_{n + 1}}}}} \right\}$
$⇒$ $S = \frac{1}{d}\left( {\frac{{nd}}{{{a_1}{a_{n + 1}}}}} \right) = \frac{n}{{{a_1}{a_{n + 1}}}}$.
Trick: Check for $n = 2$.
