Verified answer

[tex]\displaystyle\bf\\b_{n} =3x-13\\\\b_{n+1} =x-3\\\\b_{n+2}=x-5\\\\\\b_{n} \cdot b_{n+2} =b_{n+1} ^{2} \\\\\\(3x-13)\cdot(x-5)=(x-3)^{2} \\\\3x^{2} -15x-13x+65=x^{2} -6x+9\\\\3x^{2} -28x+65-x^{2} +6x-9=0\\\\2x^{2} -22x+56=0\\\\x^{2}-11x+28=0\\\\Teorema \ Vieta \ :\\\\x_{1} + x_{2} =11\\\\x_{1} \cdot x_{2} =28\\\\x_{1} =4 \ \ \ ; \ \ \ x_{2} =7\\\\1) \ x=4\\\\3x-13=3\cdot 4-13=12-13=-1\\\\x-3=4-3=1\\\\x-5=4-5=-1\\\\\boxed{-1 \ \ ; \ \ 1 \ \ ; \ \ -1}\\\\2) \ x=7[/tex]

[tex]\displaystyle\bf\\3x-13=3\cdot 7-13=21-13=8\\\\x-3=7-3=4\\\\x-5=7-5=2\\\\\boxed{8 \ \ ; \ \ 4 \ \ ; \ \ 2}\\\\Summa = \ - 1 + 1 + (-1) + 8 + 4 + 2 =13[/tex]