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"Suma a patru numere împarte consecutive este 72. Care sunt numerele?"

Răspuns :

a+a+2+a+4+a+6=72
4a+12=72
4a=72-12
4a=60
a=60:4
a=15.(primul nr)

a+2=17(al doilea nr)
a+4=19(al treilea nr)
a+6=21(al patrulea nr)

a+b+c+d=72

2k+1+2k+3+2k+5+2k+7=72

8k+16=72

8k=72-16

8k=56

k=7

a=2k+1=2*7+1=14+1=15
b=2k+3=2*7+3=14+3=17
c=2k+5=2*7+5=14+5=19
d=2k+7=2*7+7=14+7=21