Question

Sa ̆ se determine progresia aritmetic ̆a cresca ̆toare (an)n≥1 ̧stiind ca ̆ a1a2a3 = 120
a1 +a2 +a3 =15.

Answer 1

   
[tex]\displaystyle\\ \text{Termenii progresiei aritmetice sunt:}\\ a_1\\ a_2 = a_1+r\\ a_3=a_2+r=a_1+2r\\\\ \text{Scriem ecuatiile:}\\\\ \begin{cases} a_1a_2a_3 = 120\\ a_1 +a_2 +a_3 =15 \end{cases}\\\\ \begin{cases} a_1(a_1+r)(a_1+2r) = 120\\ a_1 +a_1+r +a_1+2r =15 \end{cases}\\\\ \begin{cases} a_1(a_1+r)(a_1+r+r) = 120\\ 3a_1 +3r =15~~~\Big|~:3 \end{cases}\\\\ \begin{cases} a_1(a_1+r)(a_1+r+r) = 120\\ a_1 +r =5 ~~~\text{(Inlocuim in prima ecuatie)} \end{cases}\\\\ [/tex]


[tex]\displaystyle\\ \begin{cases} a_1\cdot(5)\cdot(5+r)=120\\ a_1 +r=5 \end{cases}\\\\ \begin{cases} 5a_1(5+r)=120\\ a_1+r=5 \end{cases}\\\\ \begin{cases} 25a_1+5a_1r=120~~\Big|:5\\ a_1 +r=5 \end{cases}\\\\ \begin{cases} 5a_1+a_1r=24\\ a_1 +r =5 \end{cases}\\\\ \begin{cases} a_1(5 + r)=24\\ a_1+r=5~~\longrightarrow~\boxed{a_1=5-r} \end{cases}\\\\ (5-r)(5+r)=24\\\\ 5^2-r^2=24\\\\ r^2=25-24\\\\ r^2=1\\\\ r_{12}=\pm\sqrt{1}\\\\ \text{Progresia este crescatoare}\\\\ =\ \!\!\textgreater \ ~~\boxed{r=1}\\ a_1=5-r=5-1=4\\ \boxed{a_1=4}[/tex]


[tex]\displaystyle\\ \text{Verificare:}\\\\ r=1~~si~~a_1=4\\\\ a_1a_2a_3 =4\cdot 5\cdot 6= 120~~~Corect\\ a1 +a2 +a3 = 4+5+6 =15~~~Corect [/tex]