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}}}$