[tex]\displaystyle\bf\\q=\frac{\sqrt{3} }{6} \\\\\\S=\frac{6\cdot(\sqrt{3} +6)}{11} \\\\\\b_{3} -?\\\\\\S=\frac{b_{1} }{1-q}\\\\\\b_{1} =S\cdot(1-q)=\frac{6(\sqrt{3} +6)}{11} \cdot\Big(1-\frac{\sqrt{3} }{6} \Big)=\\\\\\=\frac{6(\sqrt{3} +6)}{11} \cdot\frac{6-\sqrt{3} }{6} =\frac{6^{2} -(\sqrt{3})^{2} }{11} =\\\\\\=\frac{36-3}{11} =\frac{33}{11} =3\\\\\\b_{3} =b_{1} \cdot q^{2} =3\cdot\Big(\frac{\sqrt{3} }{6} \Big)^{2} =3\cdot\frac{3}{36} =\frac{1}{4} =0,25\\\\\\Otvet \ : \ b_{3} =0,25[/tex]