Question

Am nevoie de ajutor la un exercitiu :
Aflati 3 nr naturale consecutive, stiind ca media lor aritmetica este egala cu

a) 90
b)105
c)2002

Multumesc ;D

Answer 1

a) x
    x+1
    x+2
$\frac{a+b+c}{3}=90$
$\frac{x+x+1+x+2}{3}=90$
$\frac{3x+3}{3}=90=\ \textgreater \ 3x+3=3*90$
3x+3=270 |-3
3x=267
x=267:3
x=89
Nr. sunt:89,90,91

b)x
   x+1
   x+2
$\frac{x+x+1+x+2}{3}=105$
$\frac{3x+3}{3}=105=\ \textgreater \ 3x+3=3*105$
3x+3=315
3x=315-3
3x=312
x=104

Nr.sunt:104,105,106

c)x
   x+1
   x+2
$\frac{x+x+1+x+2}{3}=2002$$\frac{3x+3}{3}=2002$
3x+3=3*2002
3x+3=6006 |-3
3x=6003
x=6003:3
x=2001
Nr. sunt 2001,2002,2003

Answer 2

a) a , a+1 , a+2, a+3
 4a + 6 = 90  ⇒4a= 84 ⇒  a = 21
asemenea si celelalte , b si c