Odpowiedź :
[tex]Zadanie\\\\a^2_n=a_{n-1}\cdot a_{n+1}\\\\(2x)^2=3\cdot20\\\\4x^2=60\ \ \mid:4\\\\x^2 =15\\\\x=\sqrt{15}\ \ \ \ \vee\ \ \ \ x=-\sqrt{15}[/tex]
[tex]Zadanie\\\\a^2_n=a_{n-1}\cdot a_{n+1}\\\\(2x)^2=3\cdot20\\\\4x^2=60\ \ \mid:4\\\\x^2 =15\\\\x=\sqrt{15}\ \ \ \ \vee\ \ \ \ x=-\sqrt{15}[/tex]