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સમાંતર શ્રેણીના પદો ${{\text{a}}_{\text{1}}}\text{, }{{\text{a}}_{\text{2}}}\text{, }{{\text{a}}_{\text{3}}}\text{, }......\text{ }$ લો. જો $\frac{{{a}_{1}}\,+\,\,{{a}_{2}}\,+\,....\,+\,\,{{a}_{p}}}{{{a}_{1}}\,+\,\,{{a}_{2}}\,+\,....\,+\,\,{{a}_{q}}}$ $=\,\frac{{{p}^{2}}}{{{q}^{2}}},\,p\,\,\ne \,\,q$ હોય,તો $\,\frac{{{a}_{6}}}{{{a}_{21}}}\,\,=\,\,.......$
$2/7$
$11/41$
$41/11$
$7/2$
Solution
અહીં $,\text{ }\frac{{{S}_{p}}}{{{S}_{q}}}\,\,=\,\,\frac{{{p}^{2}}}{{{q}^{2}}}\,\,\,\,$ $\Rightarrow \,\,\,\,\frac{\frac{p}{2}[2{{a}_{1}}+(p-1)d]}{\frac{q}{2}[2{{a}_{1}}+(q-1)d]}$ $\,=\,\,\frac{{{p}^{2}}}{{{q}^{2}}}\,$
$\Rightarrow \,\,\,\,\,\frac{2{{a}_{1}}+(p-1)d}{2{{a}_{1}}+(q-1)d}\,\,=\,\,\frac{p}{q}$
$\Rightarrow \,\,\,\,\frac{{{a}_{1}}+\left( \frac{p-1}{2} \right)d}{{{a}_{1}}+\left( \frac{q-1}{2} \right)d}\,\,=\,\,\frac{p}{q}\,\,\,……..(1)$
હવે, $\text{ }\frac{{{a}_{6}}}{{{a}_{21}}}\,\,=\,\,\frac{{{a}_{1}}+(6-1)d}{{{a}_{1}}+(21-1)d}\,\,\,$
$\therefore \,\,\,\,\,\frac{p-1}{2}\,\,=\,\,6-1\,\,\Rightarrow \,\,p\,\,=\,\,11\,\,$
$\therefore \,\,\,\frac{q-1}{2}\,\,=\,\,21-1\,\,\Rightarrow \,\,q\,\,=\,\,41$
પરિણામ $\text{(1) }$ આ મૂલ્યો મૂકતાં $, \frac{{{a}_{6}}}{{{a}_{21}}}\,\,=\,\,\frac{11}{41}$
