Question

Progresie geometrica: a5-a1= -1218
a4-a2= -420
a1,q=?

Answer 1

[tex] \left \{ {{a_5-a_1=-1218} \atop {a_4-a_2=-420}} \right. \Leftrightarrow \left \{ {{a_1\cdot q^4-a_1=-1218} \atop {a_1\cdot q^3-a_1\cdot q=-420}} \right. \Leftrightarrow \left \{ {{a_1(q^4-1)=-1218} \atop {a_1(q^3-q)=-420}} \right. \\ \text{Impartim relatiile:}\\ \dfrac{a_1(q_4-1)}{a_1(q^3-q)}=\dfrac{1218}{420}\\ \dfrac{(q-1)(q^3+q^2+q+1)}{(q(q-1)(q+1)}=\dfrac{203}{70}\\ \dfrac{q^3+q^2+q+1}{q^2+q}=\dfrac{203}{70}\\ 70q^3+70q^2+70q+70=203q^2+203q\\ 70q^3-133q^2-133q+70=0\\ 70q^3+70q^2-203q^2-203q+70q+70=0[/tex]
[tex]70q^2(q+1)-203q(q+1)+70(q+1)=0\\ (q+1)(70q^2-203q+70)=0|:7\\ (q+1)(10q^2-29q+70)=0\\ \text{In fine,daca descompui mai departe iti va da:}\\ (q+1)(2q-5)(5q-2)=0\\ \text{De aici avem de analizat doua cazuri(evident q nu poate fi -1):}\\ Caz\ i)q=\dfrac{5}{2}\text{ si cazul }ii)q=\dfrac{2}{5}\\ \text{Il inlocuiesti pe q in fiecare caz si il afli pe }a_1.[/tex]