Question Number 206789 by BaliramKumar last updated on 25/Apr/24
Answered by A5T last updated on 25/Apr/24
$$\sqrt[{{n}}]{{a}_{\mathrm{1}} {a}_{\mathrm{2}} ...{a}_{{n}} }={P} \\ $$$$\sqrt[{\mathrm{2}{n}}]{{a}_{{n}+\mathrm{1}} {a}_{{n}+\mathrm{2}} ...{a}_{\mathrm{3}{n}} }={Q} \\ $$$$\Rightarrow\sqrt[{\mathrm{3}}]{{PQ}^{\mathrm{2}} }=\sqrt[{\mathrm{3}{n}}]{{a}_{\mathrm{1}} {a}_{\mathrm{2}} ...{a}_{\mathrm{3}{n}} }={P}^{\frac{\mathrm{1}}{\mathrm{3}}} {Q}^{\frac{\mathrm{2}}{\mathrm{3}}} \\ $$