If {a_1},,{a_2},....,{a_{n + 1}} are in A.P., | vedclass

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Question

(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}}} +

Vedclass · Accepted Answer

$\frac{n}{{{a_1}{a_{n + 1}}}}$

Vedclass · Answer

(d) ${a_1},{a_2},{a_3},.......,{a_{n + 1}}$ are in $A.P.$

and common difference $= d$

Let $S = \frac{1}{{{a_1}{a_2}}} + \frac{1}{{{a_2}{a_3}}} + .......... + \frac{1}{{{a_n}{a_{n + 1}}}}$

$&rArr;$  $S = \frac{1}{d}\,\left\{ {\frac{d}{{{a_1}{a_2}}} + \frac{d}{{{a_2}{a_3}}} + ...... + \frac{d}{{{a_n}\,\,{a_{n + 1}}}}} \right\}$

$&rArr;$  $S = \frac{1}{d}\,\left\{ {\frac{{{a_2} - {a_1}}}{{{a_1}{a_2}}} + \frac{{{a_3} - {a_2}}}{{{a_2}{a_3}}} + ...... + \frac{{{a_{n + 1}} - {a_n}}}{{{a_n}\,\,\,{a_{n + 1}}}}} \right\}$

$&rArr;$  $S = \frac{1}{d}\left\{ {\frac{1}{{{a_1}}} - \frac{1}{{{a_2}}} + \frac{1}{{{a_2}}} - \frac{1}{{{a_3}}} + ....... + \frac{1}{{{a_n}}} - \frac{1}{{{a_{n + 1}}}}} \right\}$

$&rArr;$  $S = \frac{1}{d}\left\{ {\frac{1}{{{a_n}}} - \frac{1}{{{a_{n + 1}}}}} \right\} = \frac{1}{d}\left\{ {\frac{{{a_{n + 1}} - {a_1}}}{{{a_1}{a_{n + 1}}}}} \right\}$

$&rArr;$  $S = \frac{1}{d}\left( {\frac{{nd}}{{{a_1}{a_{n + 1}}}}} \right) = \frac{n}{{{a_1}{a_{n + 1}}}}$.

Trick: Check for $n = 2$.

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

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