Verified answer

Это геометрическая прогрессия в которой  :

[tex]\displaystyle\bf\\b_{1} =\sqrt{6} \\\\b_{2} =3\sqrt{2} \\\\\\b_{2} =b_{1} \cdot q\\\\q=\frac{b_{2} }{b_{1} } =\frac{3\sqrt{2} }{\sqrt{6} } =\frac{3\sqrt{2} }{\sqrt{2\cdot 3} } =\frac{(\sqrt{3} )^{2} \cdot\sqrt{2} }{\sqrt{2} \cdot\sqrt{3} }=\sqrt{3} \\\\\\S_{10} =\frac{b_{1} \cdot(q^{10} -1)}{q-1} =\frac{\sqrt{6} \cdot\Big[(\sqrt{3})^{10} -1\Big] }{\sqrt{3}-1 } =[/tex]

[tex]\displaystyle\bf\\=\frac{\sqrt{6} \cdot(243-1)}{\sqrt{3} -1} =\frac{242\sqrt{6} }{\sqrt{3}-1 }=\frac{242\sqrt{6} \cdot(\sqrt{3} +1)}{(\sqrt{3} -1)(\sqrt{3} +1)} =\\\\\\=\frac{242\sqrt{6} \cdot(\sqrt{3} +1)}{3-1} =121\sqrt{6} \cdot(\sqrt{3} +1)=121(3\sqrt{2} +\sqrt{6} )[/tex]