Răspuns :
a)
[tex]\div\ a_1,\ r. \\\;\\ a_1+a_2+a_3 = 27\ \ \ (1) \\\;\\ Deoarece\ \ a_1=a_2-r,\ \ a_3=a_2+r,\ \ egalitatea\ (1) \ devine\ : \\\;\\ a_2-r+a_2+a_2+r =27 \Rightarrow 3a_2=27 \Rightarrow a_2=9.\ \ \ (2) [/tex]
[tex]a_1a_2a_3=585 \Rightarrow (a_2-r)a_2(a_2+r) =585 \ \stackrel{(2)}{\Longrightarrow} (9-r)\cdot9\cdot (9+r)=585\\\;\\ Impartim\ ultima\ egalitate\ la\ 9\ si obtinem\ :\\\;\\ (9-r)(9+r) = 65\ \Rightarrow 81-r^2=65 \Rightarrow 81-65=r^2 \Rightarrow r^2 = 16\\\;\\ \Rightarrow \sqrt{r^2}=\sqrt{16} \Rightarrow |r|=4 \Rightarrow r=\pm4.[/tex]
Daca stim ratia si al doilea termen, vom determina primul termen:
[tex]a_1=a_2-r \Longrightarrow \begin{cases} a_1=9-(-4)=9+4=13\\\;\\ a_1=9-4=5\end{cases}[/tex]
Deci, avem doua progresii aritmetice:
I) 13, 9, 5, 1, ...
II) 5, 9, 13, 17, ...
[tex]\div\ a_1,\ r. \\\;\\ a_1+a_2+a_3 = 27\ \ \ (1) \\\;\\ Deoarece\ \ a_1=a_2-r,\ \ a_3=a_2+r,\ \ egalitatea\ (1) \ devine\ : \\\;\\ a_2-r+a_2+a_2+r =27 \Rightarrow 3a_2=27 \Rightarrow a_2=9.\ \ \ (2) [/tex]
[tex]a_1a_2a_3=585 \Rightarrow (a_2-r)a_2(a_2+r) =585 \ \stackrel{(2)}{\Longrightarrow} (9-r)\cdot9\cdot (9+r)=585\\\;\\ Impartim\ ultima\ egalitate\ la\ 9\ si obtinem\ :\\\;\\ (9-r)(9+r) = 65\ \Rightarrow 81-r^2=65 \Rightarrow 81-65=r^2 \Rightarrow r^2 = 16\\\;\\ \Rightarrow \sqrt{r^2}=\sqrt{16} \Rightarrow |r|=4 \Rightarrow r=\pm4.[/tex]
Daca stim ratia si al doilea termen, vom determina primul termen:
[tex]a_1=a_2-r \Longrightarrow \begin{cases} a_1=9-(-4)=9+4=13\\\;\\ a_1=9-4=5\end{cases}[/tex]
Deci, avem doua progresii aritmetice:
I) 13, 9, 5, 1, ...
II) 5, 9, 13, 17, ...
