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જો $S_1, S_2$ અને $S_3$ અનુક્રમે સમાંતર શ્રેણીના પ્રથમ $n_1, n_2$ અને $n_3$ પદોના સરવાળા દર્શાવે તો, $\frac{{{S_1}}}{{{n_1}}}\,({n_2}\, - \,{n_3})\,\, + \,\,\frac{{{S_2}}}{{{n_2}}}\,({n_3}\, - \,{n_1})\,\, + \,\,\frac{{{S_3}}}{{{n_3}}}\,({n_1}\, - \,{n_2})\,\, = ....$
$0$
$1$
$S_1S_2S_3$
$n_1n_2n_3$
Solution
અહીં $\,{{\text{S}}_{\text{1}}}\, = \,\frac{{{n_1}}}{2}\,[2a\, + \,({n_1}\, – \,1)\,d]\,$
$ \Rightarrow \,\frac{{2{S_1}}}{{{n_1}}}\, = \,2a\,({n_1}\, – \,1)d$
${S_2}\, = \,\frac{{{n_2}}}{2}\,[2a\, + \,({n_2}\, – \,1)\,d$
$ \Rightarrow \,\frac{{2{S_2}}}{{{n_2}}}\, = \,2a\, + \,({n_2}\, – \,1)\,d$
${S_3}\, = \,\frac{{{n_3}}}{2}\,[2a\, + \,({n_3}\, – \,1)d]\,$
$ \Rightarrow \,\frac{{2{S_3}}}{{{n_3}}}\, = \,2a\,({n_3} – \,1)\,d$
$\therefore \,\frac{{2{S_1}}}{{{n_1}}}\,({n_2}\, – \,{n_3})\, + \,\frac{{2{S_2}}}{{{n_2}}}\,({n_3} – {n_1})\, + \,\frac{{2{S_3}}}{{{n_3}}}\,({n_1}\, – \,{n_2})$
${\text{ = [2a + (}}{{\text{n}}_{\text{1}}}{\text{ – 1) d] (}}{{\text{n}}_{\text{2}}}{\text{ – }}{{\text{n}}_{\text{3}}}{\text{) + [2a + (}}{{\text{n}}_{\text{2}}}{\text{ – 1)d] (}}{{\text{n}}_{\text{3}}}{\text{ – }}{{\text{n}}_{\text{1}}}{\text{)}}$
${\text{ + [2a + (}}{{\text{n}}_{\text{3}}}{\text{ – 1)d] (}}{{\text{n}}_{\text{1}}}{\text{ – }}{{\text{n}}_{\text{2}}}{\text{) = 0}}$
