Question

Aflati numerele a,b,c  ce se afla in progresie geometrica ,in fiecare din cazurile:
$\left \{ {{a+b+c=26} \atop { \frac{1}{a} + \frac{1}{b}+ \frac{1}{c}= \frac{13}{18} }} \right.$
$\frac{1}{a} + \frac{1}{b}+ \frac{1}{c}= \frac{13}{18}$

$\left \{ {{a+b+c=21} \atop {abc=216}} \right.$

Answer 1

a)Presupunem ca numerele a, b si c sunt naturale.Daca aceste numere sunt in progresie geometrica atunci: $a=a, b=a\cdot q, c=a\cdot q^2$
Sistemul se scrie:
[tex]a+aq+aq^2=26\\ \frac{1}{a}+\frac{1}{aq}+\frac{1}{aq^2}= \frac{13}{18} \\ \frac{1}{a} (1+\frac{1}{q}+\frac{1}{q^2})= \frac{13}{18}\\ a(1+q+q^2)=26=>1+q+q^2= \frac{26}{a} \\ \frac{1}{a}\cdot\frac{q^2+q+1}{q^2}= \frac{13}{18} => \frac{1}{a} \cdot \frac{1}{q^2} \cdot \frac{26}{a} =\frac{13}{18}=>a\cdot q=6[/tex]
[tex]a+6+6q=26\\ a+6q=20[/tex]
Avem doua situatii:
a=2,q=3,b=6,c=18
sau
a=18.q=1/3,b=6,c=2