Question

Bardzo proszę o szybką pomoc ​

Answer 1

[tex]a_{n}\ jest\ geometryczny:\\\\\left \{ {{a_{3}=\frac{8}{17}} \atop {a_{6}= -\frac{64}{17}}} \right. \\\left \{ {{a_{1}*q^{2}=\frac{8}{17}} \atop {a_{1}*q^{5}= -\frac{64}{17}}} \right. \\\left \{ {{a_{1}=\frac{8}{17q^{2}}} \atop {a_{1}*q^{5}= -\frac{64}{17}}} \right. \\\left \{ {{a_{1}=\frac{8}{17q^{2}}} \atop {\frac{8}{17q^{2}}*q^{5}= -\frac{64}{17}}} \right. \\\left \{ {{a_{1}=\frac{8}{17q^{2}}} \atop {\frac{8}{17}*q^{3}= -\frac{64}{17}}} \right. \\[/tex]

[tex]\left \{ {{a_{1}=\frac{8}{17q^{2}}} \atop {q^{3}= -\frac{64}{17}*\frac{17}{8}}} \right. \\\left \{ {{a_{1}=\frac{8}{17q^{2}}} \atop {q^{3}= -8}} \right. \\\left \{ {{a_{1}=\frac{8}{17q^{2}}} \atop {q= -2}} \right. \\\left \{ {{a_{1}=\frac{8}{17*(-2)^{2}} \atop {q= -2}} \right. \\\left \{ {{a_{1}=\frac{8}{17*4}} \atop {q= -2}} \right. \\\left \{ {{a_{1}=\frac{8}{68}} \atop {q= -2}} \right. \\\left \{ {{a_{1}=\frac{2}{17}} \atop {q= -2}} \right.[/tex]

[tex]a_{n}=\frac{2}{17} *(-2)^{n-1}[/tex]

[tex]S_{8}=\frac{2}{17} *\frac{1-(-2)^{8}}{1-(-2)} =\frac{2}{17} *\frac{1-256}{1+2}=\frac{2}{17} *\frac{-255}{3}=-\frac{30}{3} =-10[/tex]