Wzór na n-ty wyraz ciągu geometrycznego :
[tex]a_{n} =a_{1} \cdot q^{n-1} ~~dla ~~n\geq 2\\\\a_{2} =a_{1} \cdot q\\\\a_{4} =a_{1} \cdot q^{3}\\\\a_{5} =a_{1} \cdot q^{4}\\\\podstawiam~~do~~rownan\\\\a_{1} \cdot q^{4} -a_{1} =160\\\\a_{1} \cdot (q^{4}-1)=160\\\\a_{1} \cdot (q^{2}-1)\cdot (q^{2}+1)=160\\\\\\a_{1} q^{3}-a_{1} q=48\\\\a_{1} q\cdot (q^{2} -1)=48~~\Rightarrow ~~ (q^{2} -1)=\dfrac{48}{a_{1} q } \\podstawiam~~do~~pierwszego~~rownania\\\\a_{1} \cdot \dfrac{48}{a_{1} q } \cdot (q^{2} +1)=160\\\\[/tex]
[tex]\dfrac{48(q^{2} +1)}{q }=160~~\mid \div 16\\\\\dfrac{3(q^{2} +1)}{q }=10\\\\3g^{2} +3=10q\\\\3g^{2} -10q+3=0\\\\\Delta = 100-4\cdot 3\cdot 3=100-36=64\\\sqrt{\Delta} =8\\\\q_{1} =\dfrac{10-8}{6} =\dfrac{1}{3} ~~\lor ~~q_{2} =\dfrac{10+8}{6} =3\\\\\\gdy~~q=\dfrac{1}{3} ~~to~~a_{1} ~~wynosi:\\\\a_{1} \cdot (q^{4} -1)=160\\\\a_{1} \cdot (\frac{1}{81} -1)=160\\\\-\dfrac{80}{81} \cdot a_{1} =160~~\mid \div (-\dfrac{80}{81} )\\\\a_{1} = -162\\\\\lor\\\\[/tex]
[tex]gdy~~q=3 ~~to~~a_{1} ~~wynosi:\\\\a_{1} \cdot (q^{4} -1)=160\\\\a_{1} \cdot (81 -1)=160\\\\80\cdot a_{1} =160~~\mid \div 80\\\\a_{1} =2[/tex]
[tex]Odp:~~(~~q=\dfrac{1}{3}~~\land ~~a_{1} =-162 ~~)~~\lor~~(~~q=3~~\land ~~a_{1} =2~~)[/tex]
