Предмет: Геометрия, автор: nkohska255075

Сумма первого и четвертого члена арифметической прогрессии равна 18,а шестой ее член меньше второго на 8,найдите сумму третьего и седьмого члена?

Ответы

Автор ответа: Nekit455

0

$\displaystyle\tt\left \{ {{a_1+a_4=18} \atop {a_2-8=a_6~}} \right. \left \{ {{a_1+a_4=18} \atop {a_2-a_6=8}} \right. \left \{ {{a_1+a_1+3d=18~~~~~~} \atop {a_1+d-(a_1+5d)=8}} \right. \\\\\\\left \{ {{2a_1+3d=18~~~~~~~~} \atop {a_1+d-a_1-5d=8}} \right. \left \{ {{2a_1+3d=18} \atop {-4d=8~~~~~~~}} \right. \left \{ {{2a_1+3d=18} \atop {d=-2~~~~~~~~}} \right.$

2a₁ + 3d = 18

2a₁ + 3 * (-2) = 18

2a₁ - 6 = 18

2a₁ = 18 + 6

2a₁ = 24

a₁ = 24/2

a₁ = 12

a₃ = a₁ + d(n - 1) = 12 - 2(3 - 1) = 12 - 2 * 2 = 8

a₇ = 12 - 2(7 - 1) = 12 - 2 * 6 = 0

a₃ + a₇ = 8 + 0 = 8

Ответ: 8

Автор ответа: Gentra

0

Ответ:

Ответ: По условий задачи А3+А7=8

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