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JLucas2003
@JLucas2003
December 2019
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[40 pontos]Por favor, podem me responder a essa pergunta?
Determinar a P.A. em que se verificam as relações a5+a38=125 e a18+a22=116.
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3478elc
a5+a38=125
a18+a22=116
a1 + 4r + a1+37r = 125 ==> 2a1 + 41r = 125
a1+17r+a1+21r = 116 ==> 2a1 + 38r = 116(-1)
2a1 + 41r = 125
-2a1 - 38r = -116
3r = 9 ==> r = 3
calculo de a1 :
2a1 + 41r = 125
2a1 = 125 - 41r ==> 2a1 = 125 - 41.3 ==>
2a1 = 125 - 123 ==> 2a1=2 ==> a1 = 1
(1,4,7, 10,13,16, 19,22,25, 28,31,34, 37,40,43, 46,49,52, 55,58,61,.....)
a18 + a2 ==> 52+64 = 116
1 votes
Thanks 1
Vi1984
A5 + (a5 + 33.r) = 125
2.a5 + 33r = 125 (I)
(a5 + 13.r) + (a5 + 17.r) = 116
2.a5 + 30.r = 116 (II)
(I) - (II):
3.r = 9
r = 3
2.a5 + 30.r = 116
2.a5 + 30.3 = 116
2.a5 = 116 - 90 = 26
a5 = 13
a5 = a1 + 4.r
13 = a1 + 4.3
a1 = 13-12 =1
PA:
(1,4,7,10,13,16,19,22,25,28.....)
0 votes
Thanks 1
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a5+a38=125
a18+a22=116
a1 + 4r + a1+37r = 125 ==> 2a1 + 41r = 125
a1+17r+a1+21r = 116 ==> 2a1 + 38r = 116(-1)
2a1 + 41r = 125
-2a1 - 38r = -116
3r = 9 ==> r = 3
calculo de a1 :
2a1 + 41r = 125
2a1 = 125 - 41r ==> 2a1 = 125 - 41.3 ==>
2a1 = 125 - 123 ==> 2a1=2 ==> a1 = 1
(1,4,7, 10,13,16, 19,22,25, 28,31,34, 37,40,43, 46,49,52, 55,58,61,.....)
a18 + a2 ==> 52+64 = 116
2.a5 + 33r = 125 (I)
(a5 + 13.r) + (a5 + 17.r) = 116
2.a5 + 30.r = 116 (II)
(I) - (II):
3.r = 9
r = 3
2.a5 + 30.r = 116
2.a5 + 30.3 = 116
2.a5 = 116 - 90 = 26
a5 = 13
a5 = a1 + 4.r
13 = a1 + 4.3
a1 = 13-12 =1
PA:
(1,4,7,10,13,16,19,22,25,28.....)