Question-213322

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Question Number 213322 by Spillover last updated on 02/Nov/24

Commented by Spillover last updated on 03/Nov/24

Answered by Spillover last updated on 03/Nov/24

Answered by A5T last updated on 02/Nov/24

Commented by A5T last updated on 02/Nov/24

$${R}=\mathrm{3}+\mathrm{1}+\mathrm{2}=\mathrm{6} \\ $$

Commented by Spillover last updated on 02/Nov/24

$${yes}.{great} \\ $$

Answered by mr W last updated on 02/Nov/24

((1/1)+(1/2)+(1/3)−(1/R))^2 =2((1/1^2 )+(1/2^2 )+(1/3^2 )+(1/R^2 )) (1/R^2 )+((11)/(3R))−((23)/(36))=0 ⇒(1/R)=(1/2)(−((11)/3)+((12)/3))=(1/6) ⇒R=6

$$\left(\frac{\mathrm{1}}{\mathrm{1}}+\frac{\mathrm{1}}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{3}}−\frac{\mathrm{1}}{{R}}\right)^{\mathrm{2}} =\mathrm{2}\left(\frac{\mathrm{1}}{\mathrm{1}^{\mathrm{2}} }+\frac{\mathrm{1}}{\mathrm{2}^{\mathrm{2}} }+\frac{\mathrm{1}}{\mathrm{3}^{\mathrm{2}} }+\frac{\mathrm{1}}{{R}^{\mathrm{2}} }\right) \\ $$$$\frac{\mathrm{1}}{{R}^{\mathrm{2}} }+\frac{\mathrm{11}}{\mathrm{3}{R}}−\frac{\mathrm{23}}{\mathrm{36}}=\mathrm{0} \\ $$$$\Rightarrow\frac{\mathrm{1}}{{R}}=\frac{\mathrm{1}}{\mathrm{2}}\left(−\frac{\mathrm{11}}{\mathrm{3}}+\frac{\mathrm{12}}{\mathrm{3}}\right)=\frac{\mathrm{1}}{\mathrm{6}}\:\Rightarrow{R}=\mathrm{6} \\ $$

Commented by mr W last updated on 03/Nov/24