Question

Помогите задачи по алгебре 100 баллов​

Answer 1

$\displaystyle\bf\\2)\\\\\left \{ {{a_{3}+a_{12} =2 } \atop {a_{5} +a_{14} =-14}} \right. \\\\\\\left \{ {{a_{1}+2d+a_{1} +11d=2 } \atop {a_{1} +4d+a_{1} +13d=-14}} \right. \\\\\\-\left \{ {{2a_{1}+13d=2 } \atop {2a_{1} +17d=-14}} \right. \\------------\\-4d=16\\\\d=-4\\\\2a_{1} =2-13d=2-13\cdot(-4)=2+52=54\\\\\\S_{5} =\frac{2a_{1} +4d}{2} \cdot 5=\frac{54+4\cdot (-4)}{2} \cdot 5=\frac{54-16}{2} \cdot 5=\\\\\\=\frac{38}{2} \cdot 5=19\cdot 5=95\\\\\\Otvet \ : \ d=-4 \ \ ; \ S_{5} =95$

$\displaystyle\bf\\3)\\\\\left \{ {{b_{5} -b_{3} =-48} \atop {b_{4} +b_{3}=-24 }} \right.\\\\\\\left \{ {{b_{1} \cdot q^{4} -b_{1} \cdot q^{2} =-48} \atop {b_{1} \cdot q^{3} +b_{1}\cdot q^{2} =-24 }} \right.\\\\\\:\left \{ {{b_{1} q^{2} \cdot(q^{2}-1)=-48 } \atop {b_{1} q^{2} \cdot(q+1)=-24}} \right.\\ -------------\\q-1=2\\\\q=3\\\\\\b_{1} =\frac{-24}{q^{2} \cdot(q+1)} =\frac{-24}{3^{2} \cdot(3+1)} =\frac{-24}{9\cdot 4} =-\frac{2}{3} \\\\\\Otvet \ : \ q=3 \ \ ; \ \ b_{1} =-\frac{2}{3}$