Question

Знайдіть перший член та різницю арифметичної прогресії, якщо: (впишіть коротку відповідь до 1 та 2 завдання)

Answer 1

$\displaystyle\bf\\1)\\\\\left \{ {{a_{5} + a_{12}=41 } \atop {a_{10}+a_{14} =62 }} \right. \\\\\\\left \{ {{a_{1} +4d+ a_{1}+11d=41 } \atop {a_{1}+9d+a_{1}+13d =62 }} \right.\\\\\\-\left \{ {{2a_{1}+15d=41 } \atop {2a_{1}+22d=62 }} \right. \\-----------\\-7d=-21\\\\d=3\\\\2a_{1}+22d=62\\\\a_{1}+11d=31\\\\a_{1} =31-11d=31-11\cdot 3=31-33=-2\\\\Otvet\ : \ a_{1} =-2 \ \ ; \ \ d=3$

$\displaystyle\bf\\2)\\\\\left \{ {{a_{7}+a_{13}=-104 } \atop {a_{2} \cdot a_{6}=-240}} \right. \\\\\\\left \{ {{a_{1}+6d+a_{1}+12d=-104 } \atop {(a_{1}+d) \cdot (a_{1}+5d)=-240}} \right. \\\\\\\left \{ {{2a_{1}+18d=-104 } \atop {(a_{1}+d) \cdot (a_{1}+5d)=-240}} \right. \\\\\\\left \{ {{a_{1}+9d=-52 } \atop {(a_{1}+d) \cdot (a_{1}+5d)=-240}} \right. \\\\\\\left \{ {{a_{1}=-9d-52 } \atop {(-9d-52+d) \cdot (-9d-52+5d)=-240}} \right.$

$\displaystyle\bf\\\left \{ {{a_{1}=-9d-52 } \atop {(-8d-52) \cdot (-4d-52)=-240}} \right.\\\\\\\left \{ {{a_{1}=-9d-52 } \atop {32d^{2}+416d+208d+2704 =-240}} \right. \\\\\\\left \{ {{a_{1}=-9d-52 } \atop {32d^{2}+624d+2944 =0}} \right. \\\\\\\left \{ {{a_{1}=-9d-52 } \atop {d^{2}+19,5d+92 =0}} \right. \\\\\\d^{2}+19,5d+92 =0\\\\d=(19,5)^{2} -4\cdot 92=380,25-368=12,25=3,5^{2} \\\\\\d_{1} =\frac{-19,5-3,5}{2} =-11,5\\\\\\d_{2} =\frac{-19,5+3,5}{2} =-8$

$\displaystyle\bf\\a_{1}' =-9\cdot (-11,5)-52=103,5-52=51,5\\\\a_{1}'' =-9\cdot (-8)-52=72-52=20\\\\\\Otvet \ : \ \\\\1) \ \ a_{1} =51,5 \ \ \ ; \ \ \ d=-11,5\\\\2) \ \ a_{1} =20 \ \ \ ; \ \ \ d=-8$