• Matéria: Matemática
  • Autor: grazy2araujo
  • Perguntado 8 anos atrás

interpole 15 meios aritméticos entre14 e 1116?

Respostas

respondido por: Helvio
1
Encontrar a razão da PA

an = a1 + ( n -1) . r
1116 = 14 + ( 17 -1) . r
1116 = 14 + 16r
1102 = 16r
r = 1102/16
r = 551 / 8
r = 68,875

===

an = a1 + ( n -1) . r  = an
a1 = 14 + ( 1 -1) .68,875  = 14
a2 = 14 + ( 2 -1) .
68,875   = 83
an = 14 + ( 3 -1) .
68,875   = 152 
an = 14 + ( 4 -1) .
68,875  = 221
an = 14 + ( 5 -1) .
68,875 =  290
an = 14 + ( 6 -1) .
68,875  =  358
an = 14 + ( 7 -1) .68,875  = 427
an = 14 + ( 8 -1) .68,88 = 496
an = 14 + ( 9 -1) .
68,875 = 565
an = 14 + ( 10 -1) .
68,875 =  634
an = 14 + ( 11 -1) .
68,875 = 703
an = 14 + ( 12 -1) .
68,875 = 772 
an = 14 + ( 13 -1) .
68,875 =  841
an = 14 + ( 14 -1) .68,875 =  978
an = 14 + ( 15 -1) .68,875 =  909
an = 14 + ( 16 -1) .68,875 =   1047
an = 14 + ( 17 -1) .68,875 = 1116



PA = (
14, 83, 152, 221, 290, 358, 427, 496, 565, 634, 703, 772 ,841, 909, 978, 1047,1116)

Helvio: De nada.
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