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Suma trei nr pare consecutive si dublul primului nr este un nr cuprins intre 71 si 91 sa se afle cele trei nr

Răspuns :


a+a+2+a+4+2a = 72

a+a+2+a+4+2a = 74

a+a+2+a+4+2a = 76

a+a+2+a+4+2a = 78

a+a+2+a+4+2a = 80

. . . . . . . . . . . . . . . . . .

a+a+2+a+4+2a = 90

1)

a+a+2+a+4+2a = 72

=> a+a+a+2a = 72-2-4

=> 5a = 66=> imposibil

2)

a+a+2+a+4+2a = 74

=> a+a+a+2a = 74-2-4

=> 5a = 68 => imposibil

3)

a+a+2+a+4+2a = 76

=> a+a+a+2a = 76-2-4

=> 5a = 70

=> a = 14, a+2 = 16, a+4 = 18, a = 28

4)

a+a+2+a+4+2a = 78

=> a+a+a+2a = 78-2-4

=> 5a = 72 => imposibil

5)

a+a+2+a+4+2a = 80

=> a+a+a+2a = 80-2-4

=> 5a = 74 => imposibil

. . . . . . . . . . . . . . . . . . . .

n)

a+a+2+a+4+2a = 90

=> a+a+a+2a = 90-2-4

=> 5a = 84 => imposibil