QJEEM06Ma02c1: JEE Main 2006 Mathematics - Class 11 JEEM06Ma02c1 Solution: If are in H.P., then the expression is equal to

Question

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If

{{a_1},{a_2},....{a_n}}

are in H.P., then the expression

{{a_1}\,{a_2} + \,{a_2}\,{a_3}\, + .... + {a_{n - 1}}\,{a_n}}

is equal to

Accepted Answer

{1 \over {{a_2}}} - {1 \over {{a_1}}} = {1 \over {{a_3}}} - {1 \over {{a_2}}} = .........

\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = {1 \over {{a_n}}} - {1 \over {{a_{n - 1}}}} = d

(say)

Then

{a_1}{a_2} = {{{a_1} - a{}_2} \over d},\,{a_2}{a_3} = {{{a_2} - {a_3}} \over d},

\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,......,\,{a_{n - 1}}{a_n} = {{{a_{n - 1}} - {a_n}} \over d}

\therefore

{a_1}a{}_2 + {a_2}{a_3} + ......... + {a_{n - 1}}{a_n}

= {{{a_1} - {a_2}} \over d} + {{{a_2} - {a_3}} \over d} + ......

\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, + {{{a_{n - 1}} - {a_n}} \over d}

= {1 \over a}\left[ {{a_1}} \right. - {a_2} + {a_2} - {a_3} + .......

\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left. { + {a_{n - 1}} - an} \right] = {{{a_1} - {a_n}} \over d}

Also,

{1 \over {{a_n}}} = {1 \over {{a_1}}} + \left( {n - 1} \right)d

\Rightarrow {{{a_1} - {a_n}} \over {{a_1}{a_n}}} = \left( {n - 1} \right)d

\Rightarrow {{{a_1} - {a_n}} \over d} = \left( {n - 1} \right){a_1}{a_n}

Which is the required result.

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