Question

interpole 10 meios aritméticos entre 3 e 14

Answer 1

Encontrar a razão da PA

an = a1 + ( n -1) . r

14 = 3 + ( 12 -1) . r

14 = 3 + 11r

11 = 11r

r = 1

===

an = a1 + ( n -1) . r = an

a1 = 3 + ( 1 -1) .1 = 3

a2 = 3 + ( 2 -1) .1 = 4

a3 = 3 + ( 3 -1) .1 = 5

a4 = 3 + ( 4 -1) .1 = 6

a5 = 3 + ( 5 -1) .1 =7

a6 = 3 + ( 6 -1) .1 = 8

a7 = 3 + ( 7 -1) .1 = 9

a8 = 3 + ( 8 -1) .1 = 10

a9 = 3 + ( 9 -1) .1 = 11

a10 = 3 + ( 10 -1) .1 = 12

a11 = 3 + ( 11 -1) .1 = 13

a12 = 3 + ( 12 -1) .1 = 14

PA = (3 , 4, 5 , 6, 7, 8, 9, 10, 11, 12, 13, 14 )

Answer 2

an = a1 + (n-1).r

14 = 3 + (12 - 1).r

11r = 11

r = 1

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(3,4,5,6,7,8,9,10,11,12,13,14)

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