Question

Решите Ариметическую прогрессию. (9 класс)
Даю 100 балл

Answer 1

Объяснение:

55.

$\left \{ {{a_1+a_{10}=12} \atop {a_8-a_5=4}} \right. \ \ \ \ \ \ \ \ \ \ \ \ a_{11}=?\\\left \{ {{a_1+a_1+9d=12} \atop {a_1+7d-a_1-4d=4}} \right. \ \ \ \ \ \left \{ {{2a_1+9d=12} \atop {3d=4\ |:3}} \right. \ \ \ \ \left \{ {{2a_1+9*\frac{4}{3}=12 } \atop {d=\frac{4}{3} }} \right. \ \ \ \ \ \left \{ {{2a_1+12=12} \atop {d=\frac{4}{3} }} \right. \ \ \ \ \left \{ {{a_1=0} \atop {d=\frac{4}{3} }} \right. .\\a_{11}=a_1+10d=0+10*\frac{4}{3} =\frac{40}{3}=13\frac{1}{3}.$

56.

$\left \{ {{a_5+a_{11}=-0,2} \atop {a_4-a_{10}=2,6}} \right. \ \ \ \ \ \ \ \ \ \ a_6=?\\\left \{ {{a_1+4d+a_1+10d=-0,2} \atop {a_1+3d-a_1-9d=2,6}} \right.\ \ \ \ \ \left \{ {{2a_1+14d=-0,2} \atop {-6d=2,6\ |:(-6)}} \right.\ \ \ \ \ \ \left \{ {{2a_1+14*(-\frac{13}{30} )=-0,2} \atop {d=-\frac{13}{30} }} \right. \ \ \ \ \left \{ {{2a_1-\frac{91}{15}=-\frac{1}{5} } \atop {d=-\frac{13}{30} }} \right.$

$\left \{ {{2a_1=-\frac{1}{5}+\frac{91}{15} } \atop {d=-\frac{13}{30} }} \right.\ \ \ \ \ \left \{ {{a_1=\frac{44}{15} } \atop {d=-\frac{13}{30} }} \right.\ \ \ \ \ \Rightarrow$

$a_6=a_1+5d=\frac{44}{15}+5*(-\frac{13}{30} )}=\frac{44}{15} -\frac{13}{6}=\frac{23}{30}.$

Answer 2

55. формула энного члена арифметической прогрессии такая:

an=a₁+d*(n-1)

a₁+a₁+9d=12

a₁+7d-a₁-4d=4⇒3d=4, d=4/3; 2a₁+9d=12; 2a₁=-9*4/3+12=0;

a₁₁=a₁+10d=10*4/3=40/3=13 1/3

56.  решаем аналогично.

a₁+10d+a₁+4d=-0.2⇒2a₁+14d=-0.2, а₁+7d=-0.1

a₁+3d-a₁-9d=2.6;⇒d=2.6/(-6)=-13/30;

a₁=-7*(-13/30)-0.1=91/30-3/30=88/30=44/15

а₆=a₁+5d=44/15+5*(-13/30)=(88-65)/30=23/30