Question

Insira seis meio aritméticos entre 3 e 66


1 ponto

Answer 1

Vamos là.

PA

a1 = 3

a8 = a1 + 7r = 66

3 + 7r = 66

7r = 63

r = 9

PA(3, 12, 21, 30, 39, 48, 57, 66)

Answer 2

$\large PA~ =~ ( ~3, ~12,~21, ~30, ~39, ~48,~57,~66~)$

Encontrar  a razão da PA.

$an = a1 + ( n -1) . r\\\\ 66 = 3 + ( 8 -1) . r\\\\ 66 = 3 + 7r\\\\ 63 = 7r\\\\ r = 9$

Inserir 6 meios entre 3 e 66

$an = a1 + ( n -1) . r=an\\\\a1 = 3 + ( 1 -1) .9=3\\a2 = 3 + ( 2 -1) .9=12\\a3 = 3 + ( 3 -1) .9=21\\a4 = 3 + ( 4 -1) .9=30\\a5 = 3 + ( 5 -1) .9=39\\a6 = 3 + ( 6 -1) .9=48\\a7 = 3 + ( 7 -1) .9=57\\a8 = 3 + ( 8 -1) .9=66$

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