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જો ${{\text{a}}_{\text{1}}}{\text{, }}{{\text{a}}_{\text{2}}}{\text{, }}{{\text{a}}_{\text{3}}}{\text{, }}{\text{......, }}{{\text{a}}_{\text{n}}}$ સમાંતર શ્રેણી હોય તો $\frac{1}{{{a_1}{a_2}}}\,\, + \,\,\frac{1}{{{a_2}{a_3}}}\, + \,\frac{1}{{{a_3}{a_4}}}\,\, + \,\,......\,\, + \,\frac{1}{{{a_{n - 1}}{a_n}}}\,\, = \,\,......$
$\frac{{{a_1}\,{a_2}}}{{n - 1}}$
$\frac{{n - 1}}{{{a_1} + {a_n}}}$
$\frac{{n - 1}}{{{a_1} - {a_n}}}$
$\frac{{n - 1}}{{{a_1}{a_n}}}$
Solution
આપેલ સમાંતર શ્રેણીનો સામાન્ય તફાવત $ ' d' $ લઇએ,
તો, $a_2 – a_1 = a_3 – a_2 = …….. = a_n – a_{n – 1} = d ….. (i)$
હવે, $\frac{{\text{1}}}{{{{\text{a}}_{\text{1}}}{a_2}}}\,\, + \,\,\frac{1}{{{a_2}{a_3}}}\,\, + \,\,\frac{1}{{{a_3}{a_4}}}\,\, + \,\,…….\,\, + \,\,\frac{1}{{{a_{n\, – \,1}}{a_n}}}\,\,\,\,$
$ = \,\,\frac{1}{d}\,\left[ {\frac{d}{{{a_1}{a_2}}}\,\, + \,\,\frac{d}{{{a_2}{a_3}}}\,\, + \,\,\frac{d}{{{a_3}{a_4}}}\,\, + \,\,……\,\, + \,\,\frac{d}{{{a_{n\, – \,1}}{a_n}}}} \right]$
$ = \,\,\frac{1}{d}\,\left[ {\frac{{{a_2}\, – \,\,{a_1}}}{{{a_1}{a_2}}}\,\, + \,\,\frac{{{a_3}\, – \,\,{a_2}}}{{{a_2}{a_3}}}\,\, + \,\,\frac{{{a_4}\, – \,\,{a_3}}}{{{a_4}{a_3}}}\,\, + \,\,……\,\, + \,\,\frac{{{a_n}\, – \,\,{a_{n\, – \,1}}}}{{{a_{n\, – \,1}}\,{a_n}}}} \right]$
$= \frac{1}{d}\,[ {( {\frac{1}{{{a_1}}}\,\, – \,\,\frac{1}{{{a_2}}}} )\,\, + \,\,( {\frac{1}{{{a_2}}}\,\, – \,\,\frac{1}{{{a_3}}}} )\,\, + \,\,( {\frac{1}{{{a_3}}}\,\, – \,\,\frac{1}{{{a_4}}}} )\,\, + \,\,……\,\, + \,\,( {\frac{1}{{{a_{n\, – \,1}}}}\,\, – \,\,\frac{1}{{{a_n}}}})} ]$
$ = \,\,\frac{1}{d}\,\left[ {\frac{1}{{{a_1}}}\,\, – \,\,\frac{1}{{{a_n}}}} \right]\,\, = \,\,\frac{1}{d}\,\left[ {\frac{{{a_n}\, – \,\,{a_1}}}{{{a_n}{a_1}}}} \right]\,\, = \,\,\frac{1}{d}\,\left[ {\frac{{{a_1}\, + \,\,(n\,\, – \,\,1)d\,\, – \,\,{a_1}}}{{{a_1}{a_n}}}} \right]\,\, = \,\,\frac{{n\,\, – \,\,1}}{{{a_1}{a_n}}}$
