Recordemos que la media geométrica y media aritmética de dos numeros "m" y "n" son:
[tex]\begin{array}{ccccccccccccc}\boxed{\boldsymbol{\sf{M.A.(m;n) = \dfrac{m+n}{2}}}}&&&&&&&&\boxed{\boldsymbol{\sf{M.G.(m;n) = \sqrt{m\times n}}}}\end{array}[/tex]
En el problema tenemos que:
✎ a es la media geométrica de 4 y 9
[tex]\begin{array}{c}\sf{M.G.(4;9) = \sqrt{4\times 9}}\\\\\sf{a = \sqrt{36}}\\\\\boxed{\boldsymbol{\sf{a = 6}}}\end{array}[/tex]
✎ b es la media geométrica de 2; 4 y 8
[tex]\begin{array}{c}\sf{M.G.(2;4;8)= \sqrt[3]{2\times 4\times 8}}\\\\\sf{b = \sqrt[3]{64}}\\\\\boxed{\boldsymbol{\sf{b = 4}}}\end{array}[/tex]
El problema nos pide:
[tex]\begin{array}{c}\\\sf{M.A.(a;b) = \dfrac{a+b}{2}}\\\\\sf{M.A.(a;b) = \dfrac{6+4}{2}}\\\\\sf{M.A.(a;b) = \dfrac{10}{2}}\\\\\boxed{\boxed{\boldsymbol{\red{\sf{M.A.(a;b) = 5}}}}}\end{array}[/tex]
[tex]\boxed{\sf{{R}}\quad\raisebox{10pt}{$\sf{\red{O}}$}\!\!\!\!\raisebox{-10pt}{$\sf{\red{O}}$}\quad\raisebox{15pt}{$\sf{{G}}$}\!\!\!\!\raisebox{-15pt}{$\sf{{G}}$}\quad\raisebox{15pt}{$\sf{\red{H}}$}\!\!\!\!\raisebox{-15pt}{$\sf{\red{H}}$}\quad\raisebox{10pt}{$\sf{{E}}$}\!\!\!\!\raisebox{-10pt}{$\sf{{E}}$}\quad\sf{\red{R}}}\hspace{-64.5pt}\rule{10pt}{.2ex}\:\rule{3pt}{1ex}\rule{3pt}{1.5ex}\rule{3pt}{2ex}\rule{3pt}{1.5ex}\rule{3pt}{1ex}\:\rule{10pt}{.2ex}[/tex]
