Найдите сумму первых 6 членов геометрической прогрессии если b6=25 , b8=9.

[latex]b_6=25,\; b_8=9\\\\ \frac{b_8}{b_6} = \frac{b_1q^7}{b_1q^5} =q^2\; \; \to \; \; q^2= \frac{9}{25} =(\frac{3}{5})^2\; \; \to \\\\q=\pm \frac{3}{5}\\\\b_1=\frac{b_6}{q^5}=\frac{25}{\pm (\frac{3}{5})^5}=\pm \frac{5^7}{3^5}=\pm \frac{3125}{243}\\\\Esli\; b_1=-\frac{3125}{243},\; to\; q=-\frac{3}{5}.\\\\Esli\; b_1=+\frac{3125}{243},\; to\; q=+\frac{3}{5}\\\\S_6= \frac{b_1-b_6q}{1-q}\\\\1)S_6 = \frac{\frac{3125}{243}-25\cdot \frac{3}{5}}{1-\frac{3}{5}} = \frac{5\cdot 3125-25\cdot 3\cdot 243}{243\cdot 2} =[/latex] [latex]= \frac{-2600}{486} =-5 \frac{170}{243} \\\\2)\; \; S_6= \frac{-\frac{3125}{243}+\frac{3}{5}\cdot 25}{1+\frac{3}{5}}= \frac{2600}{1944}=1\frac{656}{1944} [/latex]