Question

Arătați că numerele $\sqrt{11} - \sqrt{5}$, $\sqrt{6}$ și $\sqrt{11} + \sqrt{5}$ sunt termeni consecutivi ai unei progresii geometrice.

Answer 1

 

$\displaystyle\bf\\Metoda~1:\\Daca~~a,~b,~c~sunt~in~progresie~geometrica~atunci:~~\frac{b}{a}=\frac{c}{b}\\\\\frac{\sqrt{6}}{\sqrt{11}-\sqrt{5}}=\frac{\sqrt{11}+\sqrt{5}}{\sqrt{6}}\\\\\Big(\sqrt{11}+\sqrt{5}\Big)\Big(\sqrt{11}-\sqrt{5}\Big)=\Big(\sqrt{6}\Big)^2\\\\\Big(\sqrt{11}\Big)^2-\Big(\sqrt{5}\Big)^2=\Big(\sqrt{6}\Big)^2\\\\11-5=6\\\\6=6\\\\\implies~~\sqrt{11}-\sqrt{5},~\sqrt{6},~\sqrt{11}+\sqrt{6} = pg$

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$\displaystyle\bf\\Metoda~2:\\Daca~~a,~b,~c~sunt~in~progresie~geometrica~atunci:~~\sqrt{a\times c}=b\\\\\sqrt{\Big(\sqrt{11}+\sqrt{5}\Big)\Big(\sqrt{11}-\sqrt{5}\Big)}=\\\\=\sqrt{\Big(\sqrt{11}\Big)^2-\Big(\sqrt{5}\Big)^2}=\\\\=\sqrt{11-5}=\sqrt{6}\\\\\implies~~\sqrt{11}-\sqrt{5},~\sqrt{6},~\sqrt{11}+\sqrt{6} = pg$