Question

1.A)suma a trei numere naturale consecutive pare este 252.Determinati numerele.

B)suma a cinci numere consecutive este 75.Determinati numerele.

C)suma a patru numere impare consecutive este 72.Determinati-le

2.Suma a trei numere este 695.Diferenta dintre primul numar si al doilea este 256,iar diferenta dintre primul si al treilea este de 192.aflati cele tri numere.

Answer 1

1.A)2k+2+2k+4+2k+6=252

6k+12=252

6k=252-12

6k=240

k=40

2k+2=2*40+2=80+2=82
2k+4=2*40+4=80+4=84
2k+6=2*40+6=80+6=86
B)
n+n+1+n+2+n+3+n+4 = 75
5n+10=75
5n = 75-10
5n = 65 / :5
n= 13
C)
n + ( n+2) + (n+4) + (n+6) = 72
4n+ 12 = 72 ; 4n = 72-12 ; 4n=60 /:4 ; n=15. 15+2 = 17 ; 15+4 = 19 ; 15+6 = 21. V = 15+17+19+21 = 65 ( nu am putut sa le scriu una sub alta...)
2)a+b+c=695
a-b=256⇒a=b+256
a-c=192⇒a=c+192
b+256-192=c⇒c-b=256-192⇒c=+64

b+256+b+64=695
3b+320=695⇒3b=375
b=375:3=125
a=125+256=381
c=125+64=189

Answer 2

a+b+c=252
b=a+2
c=a+4
a+a+2+a+4=252->3a=252-6->3a=246->a=246/3=82
a=82  
b=84
c=86


a+b+c+d+e=75
b=a+1
c=a+2
d=a+3
e=a+4
a+a+1+a+2+a+3+a+4=75->5a+10=75->5a=75-10=65->a=65/5->a=13
b=14
c=15
d=16
e=17

a+b+c+d=72
b=a+2
c=a+4
d=a+6
a+a+2+a+4+a+6=72->4a+12=72->4a=72-12=60->a=60/4->a=15
b=17
c=19
d=21

a+b+c=695
a-b=256->b=a-256
a-c=192->c=a-192
a+(a-256)+(a-192)=695->3a-256-192=695->3a=695+192+256=1143->a=1143/3->a=381
b=381-256=125
c=381-192=189