Question

6. Найдите первый член а1 арифмеической прогрессии и различие d, когда
а){a1+3d=-6
{a1-4d=8
b){a6+a8=82
{a5-a3=12
c){a9+a3=29
{a2*a7=9​

Answer 1

Ответ:

$a)a_1=0,d=-2\\b)a_1=5,d=6\\c)a_1=3,d=3,5\\ a_1=86,375 d=-14,375$

Объяснение:

$a)\\\left \{ {{a_1+3d=-6} \atop {a_1-4d=8}} \right. \Rightarrow \left \{ {{7d=-14} \atop {a_1-4d=8}} \right. \Rightarrow \left \{ {{d=-2} \atop {a_1-4d=8}} \right. \Rightarrow \left \{ {{d=-2} \atop {a_1-4*(-2)=8}} \right. \Rightarrow \left \{ {{d=-2} \atop {a_1+8=8}} \right. \Rightarrow \left \{ {{d=-2} \atop {a_1=0}} \right.$

$b)\\\left \{ {{a_6+a_8=82} \atop {a_5-a_3=12}} \right. \Rightarrow \left \{ {{a_1+5d+a_1 +7d=82} \atop {a_1+4d-a_1-2d=12}} \right. \Rightarrow \left \{ {{2a_1+12d=82} \atop {2d=12}} \right. \Rightarrow \left \{ {{2a_1+12d=82} \atop {d=6}} \right. \Rightarrow\left \{ {{2a_1+72=82} \atop {d=6}} \right. \Rightarrow \left \{ {{2a_1=10} \atop {d=6}} \right. \Rightarrow \left \{ {{a_1=5} \atop {d=6}} \right.$

$c)\\\left \{ {{a_9+a_3=29} \atop {a_2*a_7=9}} \right. \Rightarrow \left \{ {{a_1+8d+a_1+2d=29} \atop {(a_1+d)*(a_1+6d)=9}} \right. \Rightarrow \left \{ {{2a_1+10d=29} \atop {(a_1+d)*(a_1+6d)=9}} \right. \Rightarrow \left \{ {{a_1=\frac{29-10d}{2}} \atop {(a_1+d)*(a_1+6d)=9}} \right. \Rightarrow$

$\left \{ {{a_1=\frac{29-10d}{2}} \atop {(\frac{29-10d}{2}+d)*(\frac{29-10d}{2}+6d)=9}} \right. \Rightarrow \left \{ {{a_1=\frac{29-10d}{2}} \atop {(14,5-4d)*(14,5+d)=9}} \right. \Rightarrow \left \{ {{a_1=\frac{29-10d}{2}} \atop {14,5^2-3*14,5d-4d^2=9}} \right. \Rightarrow$

$\left \{ {{a_1=\frac{29-10d}{2}} \atop {29^2-3*58d-16d^2=36}} \right. \Rightarrow \left \{ {{a_1=\frac{29-10d}{2}} \atop {16d^2+174d-805=0}} \right.$

$D=174^2-4*16*(-805)=81796=286^2\\d_{1}=\frac{-174+286}{32}==\frac{112}{32}=3,5 \Rightarrow a_1=\frac{29-35}{2}=-3\\d_2=\frac{-174-286}{32}=\frac{-460}{32}=-14,375\Rightarrow a_1=\frac{29+143,75}{2}=86,375$