Question

Se dau 3 numere naturale .Media aritmetica primelor doua este 14 .Media aritmetica ultimelor 2 este 20 .Ultimul numar e de 4 ori mai mare decat al doilea .Calculati .​

Answer 1

Avem a, b si c.

(a+b) : 2 = 14 => a+b = 28 (1)

(b+c) : 2 = 20 => b+c = 40 (2)

c=4b

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In (2),

b+c = 40

b+4b = 40

5b = 40

=> b = 40/5 = 8

b = 8

_____________________

=> c = 40-b

c = 40 - 8 = 32

c = 32

_____________________

In (1),

a+b = 28

=> a = 28-b

a = 28-8

a = 20

Answer 2

Răspuns:

(a+b)/2=14=>a+b=14·2=>a+b=28

(b+c)/2=20=>b+c=20·2=>b+c=40

c=4b

b+4b=40=>b=40:5=>b=8

c=4·8=32

a+b=28=>a+8=28=>a=28-8=>a=20