\(\displaystyle{ a_{1}=b_{1}}\)
\(\displaystyle{ a_{2}=b_{4}}\)
\(\displaystyle{ a_{3}=b_{5}}\)
\(\displaystyle{ \frac{b_{5}}{b_{4}} = \frac{b_{4}}{b_{1}}}\)
\(\displaystyle{ b_{5} - b_{4} = b_{4} - b_{3} = r}\)
\(\displaystyle{ b_{2} = b_{1} + r}\)
\(\displaystyle{ b_{3} = b_{1} + 2r}\)
\(\displaystyle{ \frac{b_{1} + 4r}{b_{1} + 3r} = \frac{b_{1} + 3r}{b_{1}} = q}\)
\(\displaystyle{ \frac{b_{1} + 4r}{b_{1} + 3r} = q}\)
\(\displaystyle{ { \{ }}\)
\(\displaystyle{ \frac{b_{1} + 3r}{b_{1}} = q}\)
(te dwa w klamre ;p nie wiem jak taka duza zrobic ;])
i z tego (2)
\(\displaystyle{ 1+ \frac{3r}{b_{1}}=q}\)
\(\displaystyle{ \frac{3r}{1-q}=b_{1}}\)
\(\displaystyle{ \frac{\frac{3r}{1-q} +4r} {\frac{3r}{1-q} +3r} = q / *(1-q)}\)
\(\displaystyle{ \frac{3r +4r(1-q)} {3r +3r(1-q)} = q}\)
\(\displaystyle{ \frac{3 +4-4q} {3+3-3q} = q - q^{2}}\)
\(\displaystyle{ 7-4q-(q-q^{2})(6-3q)=0}\)
\(\displaystyle{ 7-4q-(6q-6q^{2}-3q^{2}+3q^{3})=0}\)
\(\displaystyle{ -3q^{3}+9q^{2}-10q+7=0}\)
|q|