Question

СТРОЧНО!!!
ДАМ 20 БАЛОВ
а4+а7=15 ; а9-а2=а10 ;
а20=?​

Answer 1

$\left \{ {{a_{4}+a_{7}=15} \atop {a_{9} -a_{2}=a_{10} }} \right.\\\\\left \{ {{a_{1}+3d+a_{1}+6d=15 } \atop {a_{1}+8d-a_{1}-d-a_{1}-9d=0}} \right.\\\\\left \{ {{2a_{1}+9d=15 } \atop {-a_{1}-2d=0 }} \right.\\\\\left \{ {{a_{1}=-2d } \atop {2*(-2d)+9d=15}} \right.\\\\\left \{ {{a_{1}=-2d } \atop {5d=15}} \right.\\\\\left \{ {{a_{1}=-6 } \atop {d=3}} \right.\\\\a_{20}=a_{1}+19d=-6+19*3=-6+57=51\\\\\boxed{a_{20}=51}$

Answer 2

Объяснение:

$\left \{ {{a_4+a_7=15} \atop {a_9-a_2=a_{10}}} \right. \ \ \ \ \ \left \{ {{a_1+3d+a_1+6d=15} \atop {a_1+8d-(a_1+d)=a_1+9d}} \right. \ \ \ \ \left \{ {{2a_1+9d=15} \atop {a_1+8d-a_1-d=a_1+9d}} \right. \\\left \{ {{2a_1+9d=15} \atop {a_1+2d=0\ |*2}} \right. \left \{ {{2a_1+9d=15} \atop {2a_1+4d=0}} \right..$

Вычитаем из первого уравнения второе:

$5d=15\ |:5\\d=3.\ \ \ \ \Rightarrow\\a_1+2*3=0\\a_1+6=0\\a_1=-6.\\a_{20}=a_1+19d=-6+19*3=-6+57=51.$

Ответ: a₂₀=51.