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Question-206789




Question Number 206789 by BaliramKumar last updated on 25/Apr/24
Answered by A5T last updated on 25/Apr/24
((a_1 a_2 ...a_n ))^(1/n) =P  ((a_(n+1) a_(n+2) ...a_(3n) ))^(1/(2n)) =Q  ⇒((PQ^2 ))^(1/3) =((a_1 a_2 ...a_(3n) ))^(1/(3n)) =P^(1/3) Q^(2/3)
$$\sqrt[{{n}}]{{a}_{\mathrm{1}} {a}_{\mathrm{2}} …{a}_{{n}} }={P} \\ $$$$\sqrt[{\mathrm{2}{n}}]{{a}_{{n}+\mathrm{1}} {a}_{{n}+\mathrm{2}} …{a}_{\mathrm{3}{n}} }={Q} \\ $$$$\Rightarrow\sqrt[{\mathrm{3}}]{{PQ}^{\mathrm{2}} }=\sqrt[{\mathrm{3}{n}}]{{a}_{\mathrm{1}} {a}_{\mathrm{2}} …{a}_{\mathrm{3}{n}} }={P}^{\frac{\mathrm{1}}{\mathrm{3}}} {Q}^{\frac{\mathrm{2}}{\mathrm{3}}} \\ $$

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