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જો સમાંતર શ્રેણી નું $m$ મું પદ $1/n$ અને $n$ મું પદ $1/m$ હોય તો $mn$ પદોનો સરવાળો ......થાય.
$mn + 1$
$\frac{{\rm{1}}}{{\rm{2}}}{\rm{(2mn}} + {\rm{1)}}$
$\frac{{\rm{1}}}{{\rm{2}}}{\rm{(mn}} + {\rm{1)}}$
$2mn + 1$
Solution
$\frac{{\text{1}}}{{\text{n}}} = {\text{a}} + {\text{(m}} – {\text{1)d}}\,\,\,\,\,\,\,…….{\text{(1)}}\,$
$\frac{{\text{1}}}{{\text{m}}} = {\text{a}} + {\text{(n}} – {\text{1)d }}……{\text{(2)}}$
સમીકરણ (1) અને (2) ઉકેલતાં $ \Rightarrow \,\,\,{\text{a}} = \frac{{\text{1}}}{{{\text{mn}}}},\,\,\,\,{\text{d}} = \frac{{\text{1}}}{{{\text{mn}}}}$
${{\text{S}}_{{\text{mn}}}} = \frac{{{\text{mn}}}}{{\text{2}}}\left[ {\frac{{\text{2}}}{{{\text{mn}}}} + {\text{(mn}} – {\text{1)}}\frac{{\text{1}}}{{{\text{mn}}}}} \right] = \frac{{\text{1}}}{{\text{2}}}{\text{(mn}} + {\text{1)}}$
