Question

Media aritmetica a 3 numere este 21 , iar media aritmetica  a primelor doua numere este 25 . Calculati numerele , stiind ca al treilea numar este cu 7 mai mic decat dublul primului numar . 

Answer 1

[tex]m_a= \frac{a+b+c}{3} \ \textless \ =\ \textgreater \ 21= \frac{a+b+c}{3} \\\\ m_a= \frac{a+b}{2} \ \textless \ =\ \textgreater \ 25= \frac{a+b}{2}=\ \textgreater \ a+b=25*2=\ \textgreater \ a+b=50\\\\ c=2*a-7\\ ---------------\\ =\ \textgreater \ 21= \frac{50+c}{3} =\ \textgreater \ 50+c=63=\ \textgreater \ c=63-50=\ \textgreater \ c=13\\\\ =\ \textgreater \ 13=2a-7=\ \textgreater \ 2a=20=\ \textgreater \ a=20:2=\ \textgreater \ a=10\\\\ 25= \frac{10+b}{2} =\ \textgreater \ 10+b=50=\ \textgreater \ b=40\\\\ ---------------\\ Proba: \frac{40+10+13}{3} =\ \textgreater \ 63:3=\ \textgreater \ 21(corect)\\\\ \frac{40+10}{2}=50:2=\ \textgreater \ 25(corect)\\\\ 13=2*10-7=\ \textgreater \ 13=20-7=\ \textgreater \ 13=13(corect) [/tex]