Interpole cinco meios aritméticos entre 12 e 44 me ajudem por favor
Soluções para a tarefa
Respondido por
1
a₁ = 12
a₇ = 44 ------ 44 = 12 + (7 - 1) . r ---- 44 - 12 = 6r ---- 6r = 32 ----- r = 32/6
r = 16/3
a₂ = 12 + (2 - 1) . 16/3 ------- a₂ = 12 + 1 . 16/3 ---- a₂ = 12 + 16/3
a₂ = (36 + 16)/3 ----- a₂ = 52/3
a₃ = 12 + (3 - 1) . 16/3 ------- a₃ = 12 + 2 . 16/3 ---- a₃ = 12 + 32/3
a₃ = (36 + 32)/3 ---- a₃ = 68/3
a₄ = 12 + (4 - 1) . 16/3 ---- a₄ = 12 + (3) . 16/3 ---- a₄ = 12 + 16 ---- a₄ = 28
a₅ = 12 + (5 - 1) . 16/3 ---- a₅ = 12 + (4) . 16/3 ---- a₅ = 12 + 64/3
a₅ = (36 + 64)/3 ----- a₅ = 100/3
a₆ = 12 + (6 - 1) . 16/3 ----- a₆ = 12 + (5) . 16/3 ----- a₆ = 12 + 80/3
a₆ = (36 + 80)/3 ----- a₆ = 116/3
(12, 52/3, 68/3, 28, 100/3, 116/3, 44)
a₇ = 44 ------ 44 = 12 + (7 - 1) . r ---- 44 - 12 = 6r ---- 6r = 32 ----- r = 32/6
r = 16/3
a₂ = 12 + (2 - 1) . 16/3 ------- a₂ = 12 + 1 . 16/3 ---- a₂ = 12 + 16/3
a₂ = (36 + 16)/3 ----- a₂ = 52/3
a₃ = 12 + (3 - 1) . 16/3 ------- a₃ = 12 + 2 . 16/3 ---- a₃ = 12 + 32/3
a₃ = (36 + 32)/3 ---- a₃ = 68/3
a₄ = 12 + (4 - 1) . 16/3 ---- a₄ = 12 + (3) . 16/3 ---- a₄ = 12 + 16 ---- a₄ = 28
a₅ = 12 + (5 - 1) . 16/3 ---- a₅ = 12 + (4) . 16/3 ---- a₅ = 12 + 64/3
a₅ = (36 + 64)/3 ----- a₅ = 100/3
a₆ = 12 + (6 - 1) . 16/3 ----- a₆ = 12 + (5) . 16/3 ----- a₆ = 12 + 80/3
a₆ = (36 + 80)/3 ----- a₆ = 116/3
(12, 52/3, 68/3, 28, 100/3, 116/3, 44)
Perguntas interessantes