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Question Number 31717 by mondodotto@gmail.com last updated on 13/Mar/18

Answered by Giannibo last updated on 13/Mar/18

(a,b,c) c=λb & b=λa⇒c=λ^2 a abc=1000⇒λ^3 a^3 =1000 ⇒ λ=((10)/a)(2) 7+c=6+b+ω⇒7+λ^2 a=6+λa+ω⇒^((1)) 1+λ^2 a=2λa+6−a⇒^((2)) 1+((100)/a)=26−a⇒a^2 −25a+100⇒a=((25±15)/2)=20 or 5(3) 6+b=a+ω⇒6+λa−a=ω (1) If a=20 ⇒ λ=0.5 ⇒ b=10 ⇒ c=5 If a=5 ⇒ λ=2 ⇒ b=10 ⇒ c=20

$$ \\ $$$$ \\ $$$$\left({a},{b},{c}\right) \\ $$$${c}=\lambda{b}\:\&\:{b}=\lambda{a}\Rightarrow{c}=\lambda^{\mathrm{2}} {a} \\ $$$${abc}=\mathrm{1000}\Rightarrow\lambda^{\mathrm{3}} {a}^{\mathrm{3}} =\mathrm{1000}\:\Rightarrow\:\lambda=\frac{\mathrm{10}}{{a}}\left(\mathrm{2}\right) \\ $$$$\mathrm{7}+{c}=\mathrm{6}+{b}+\omega\Rightarrow\mathrm{7}+\lambda^{\mathrm{2}} {a}=\mathrm{6}+\lambda{a}+\omega\overset{\left(\mathrm{1}\right)} {\Rightarrow}\mathrm{1}+\lambda^{\mathrm{2}} {a}=\mathrm{2}\lambda{a}+\mathrm{6}−{a}\overset{\left(\mathrm{2}\right)} {\Rightarrow}\mathrm{1}+\frac{\mathrm{100}}{{a}}=\mathrm{26}−{a}\Rightarrow{a}^{\mathrm{2}} −\mathrm{25}{a}+\mathrm{100}\Rightarrow{a}=\frac{\mathrm{25}\pm\mathrm{15}}{\mathrm{2}}=\mathrm{20}\:\mathrm{or}\:\mathrm{5}\left(\mathrm{3}\right) \\ $$$$\mathrm{6}+{b}={a}+\omega\Rightarrow\mathrm{6}+\lambda{a}−{a}=\omega\:\left(\mathrm{1}\right) \\ $$$$\mathrm{If}\:{a}=\mathrm{20}\:\Rightarrow\:\lambda=\mathrm{0}.\mathrm{5}\:\Rightarrow\:{b}=\mathrm{10}\:\Rightarrow\:{c}=\mathrm{5} \\ $$$$\mathrm{If}\:{a}=\mathrm{5}\:\Rightarrow\:\lambda=\mathrm{2}\:\Rightarrow\:{b}=\mathrm{10}\:\Rightarrow\:{c}=\mathrm{20} \\ $$

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