[tex]\it Fie\ a,\ aq,\ aq^2\ cele\ trei\ numere,\ ale\ progresiei\ geometrice .\\ \\ a + aq + aq^2 = 26\ \ \ \ \ (1)\\ \\ a+1,\ aq+6,\ aq^2+3\ formeaz\breve{a}\ o\ progresie\ aritmetic\breve{a}.\\ \\ Suma\ lor\ este:\\ \\ S_3=a+1+aq+6+aq^2+3=(a+aq+aq^2)+10\ \stackrel{(1)}{=}\ 26+10=36\ \ \ \ \ (2)[/tex]
[tex]\it Dar,\ conform\ formulei\ de\ la\ progresii\ aritmetice, avem:\\ \\ S_3=\dfrac{(a+1+aq^2+3)\cdot3}{2}=\dfrac{(a+aq^2+4)\cdot3}{2}\ \ \ \ \ (3)\\ \\ (2),\ (3) \Rightarrow \dfrac{(a+aq^2+4)\cdot3}{2}=36|_{\cdot\dfrac{2}{3}}\Rightarrow a+aq^2+4=24|_{-4} \Rightarrow \\ \\ \Rightarrow a+aq^2=20\ \ \ \ \ (4)[/tex]
[tex]\it (1),\ (4)\Rightarrow 20+aq=26|_{-20} \Rightarrow aq=6\Rightarrow a=\dfrac{6}{q}\ \ \ \ \ (5)[/tex]
[tex]\it (1),\ (5)\Rightarrow \dfrac{6}{q}+6+6q=26|_{-6}\Rightarrow \dfrac{6}{q}+6q=20|_{\cdot q}\Rightarrow 6+6q^2=20q\Rightarrow\\ \\ \Rightarrow6q^2-20q+6=0|_{:2}\Rightarrow3q^2-10q+3=0\Rightarrow3q^2-9q-q+3=0\Rightarrow\\ \\ 3q(q-3)-(q-3)=0\Rightarrow (q-3)(3q-1)=0\Rightarrow q_1=\dfrac{1}{3},\ \ q_2=3[/tex]
[tex]\it q=\dfrac{1}{3}\ \stackrel{(5)}{\Longrightarrow}\ a=\dfrac{6}{\dfrac{1}{3}}=6\cdot3=18\Rightarrow progresia\ geometric\breve{a}:\\ \\ 18,\ \ 6,\ \ 2,\ ...\ \ descresc\breve{a}toare\ (nu\ convine).\\ \\ q=3\ \stackrel{(5)}{\Longrightarrow}\ a=\dfrac{6}{3}=2\Rightarrow progresia\ geometric\breve{a}:\\ \\ 2,\ \ 6,\ \ 18,\ ...\ \ cresc\breve{a}toare\ (\ convine !\ ).[/tex]
Prin urmare, numerele cerute sunt: 2, 6, 18.
