Предмет: Алгебра, автор: Олька124

Дана геометрическая прогрессия 125,25;....Найдите сумму первых пяти членов этой прогрессии

Ответы

Автор ответа: tslayer67
0
q=25/125=1/5
b5= b1q^4=125*(1/5)^4=125/625=1/5
S5=-624/25*5/4=156/5=31,2
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