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જો ${\text{r}}\,\, > \,\,{\text{1}}$ અને ${\text{x}}\, = \,\,{\text{a}}\, + \,\frac{a}{r}\, + \,\frac{a}{{{r^2}}}\, + \,..\,\,\infty ,\,\,y\, = \,b\, - \,\frac{b}{r}\, + \,\frac{b}{{{r^2}}} - \,..\,\,\,\infty $ અને ${\text{z}}\,\, = \,\,{\text{c}}\, + \,\frac{{\text{c}}}{{{{\text{r}}^{\text{2}}}}}\, + \,\frac{c}{{{r^4}}}\, + \,\,\,\infty ,\,$ હોય, તો $\frac{{{\text{xy}}}}{{\text{z}}}\,\, = \,...$
$ab/c$
$ac/b$
$bc/a$
આપેલ પૈકી એક પણ નહિ
Solution
$\,\,x\,\, = \,\,\,\frac{a}{{1\,\, – \,\,\frac{1}{r}}}\,\,;\,\,\,\,y\,\, = \,\,\frac{b}{{1\,\, + \,\,\frac{1}{r}}}\,\,;\,\,\,z\,\, = \,\,\frac{c}{{1\,\, – \,\,\frac{1}{{{r^2}}}}}$
$1\,\, – \,\,\frac{1}{r}\,\, = \,\,\frac{a}{x}\,\,;\,\,\,\,1\,\, + \,\,\frac{1}{r}\,\, = \,\,\frac{b}{y}\,\,\,\,\, \Rightarrow \,\,\,\,z\,\, = \,\,\frac{c}{{\frac{a}{x}\,.\,\frac{b}{y}}}$
$\frac{z}{{xy}}\,\, = \,\,\frac{c}{{ab}}\,\,\,\,\,\,\, \Rightarrow \,\,\frac{{xy}}{z}\,\, = \,\,\frac{{ab}}{c}$
