арифметические прогрессия (аn) дана условиями: a1=-12, an=an-1+12(при n>1). найдите

арифметические прогрессия (аn) дана условиями: a1=-12, an=an-1+12(при n>1). найдите сумму первых 7 её членов. 168 должно получиться
- Предмет:
  Математика
- Автор:
  leonardomoreno
- 6 лет назад

Ответы 1

$a2 = a1 + 12 = - 12 + 12 = 0 \\ a3 = a2 + 12 = 0 + 12 = 12 \\ a4 = a3 + 12 = 12 + 12 = 24 \\ a5 = a4 + 12 = 24 + 12 = 36 \\ a6 = a5 + 12 = 36 + 12 = 48 \\ a7 = a6 + 12 = 48 + 12 = 60 \\ a1 + a2 + a3 + a4 + a5 + a6 + a7 = - 12 + 0 + 12 + 24 + 36 + 48 + 60 = 168$
- Автор:
  marquis
- 6 лет назад
- 0
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арифметические прогрессия (аn) дана условиями: a1=-12, an=an-1+12(при n>1). найдите сумму первых 7 её членов. 168 должно получиться

Ответы 1

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