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1) Пусть а - первый член геометрической прогрессии
2) Тогда третий член прогрессии: а·q²
3) Значит, сумма первого и третьего членов будет (а+а·q²) или а(1+q²)=10
4) Второй член прогрессии выразится как а·q
5) Четвёртый член выразится как а·q³
6) Тогда сумма второго и четвёртого будет а·q+а·q³ или а(q+q³)=30
7) Разделите выражение (3) на выражение (6). Точнее, левую часть на левую, а правую на правую. Вы должны получить :
(1+q²)/(q+q³)=(1/3) или 3(1+q²)=(q+q³) или 3(1+q²)=q(1+q²) ⇒q=3
8) По условию известно, что сумма первого и третьего равна 10:
а(1+q²)=10 или а(1+3²)=10 ⇒ 10·а=10 ⇒ а=1( это ответ)
УДАЧИ!
Пошаговое объяснение:
а) х + 100 = 184 : 2 х + 100 = 484 : 4
х = 92 - 100 х = 121 - 100
х = -8 х = 21
-8 < 21
б) х * 2 = 109 + 113 х * 2 = 109 + 313
х = 222 : 2 х = 422 : 2
х = 111 х = 211
111 < 211