Question

35балов помогите срочно

1-Задание
Найти число членов арефмитической прогрессии, если а3-а1=12; а2+а4=18.Sn=105




2-Задание
Найдите сумму бесконечно убывающей прогрессии 49;7;1;/7;.....​

Answer 1

$\displaystyle\bf\\\left \{ {a_{3} -a_{1} =12} \atop {a_{2} +a_{4} =18}} \right. \\\\\\\left \{ {{a_{1} +2d-a_{1} =12} \atop {a_{1} +d+a_{1} +3d=18}} \right. \\\\\\\left \{ {{2d=12} \atop {2a_{1} +4d=18}} \right. \\\\\\\left \{ {{d=6} \atop {a_{1} +2d=9}} \right. \\\\\\\left \{ {{d=6} \atop {a_{1}=9-2d }} \right. \\\\\\\left \{ {{d=6} \atop {a_{1} =9-12=-3}} \right. \\\\\\S_{n} =\frac{2a_{1} +d\cdot(n-1)}{2} \cdot n\\\\\\105=\frac{2\cdot(-3)+6\cdot(n-1)}{2} \cdot n$

$\displaystyle\bf\\210=(-6+6n-6)\cdot n\\\\210=(6n-12)\cdot n\\\\6n^{2} -12n-210=0\\\\n^{2} -2n-35=0\\\\Teorema \ Vieta:\\\\n_{1} =7\\\\n_{2} =-5<0-neyd\\\\Otvet:7\\\\\\2)\\\\b_{1} =49\\\\b_{2} =7\\\\b_{2} =b_{1}\cdot q\\\\q=b_{2} :b_{1}=7:49=\frac{1}{7}\\\\\\S=\frac{b_{1} }{1-q} =\frac{49}{1-\frac{1}{7} } =\frac{49}{\frac{6}{7} } =57\frac{1}{6} \\\\\\Otvet:S=57\frac{1}{6}$