Răspuns:
Explicație pas cu pas:
[tex]2+S_{2}=0 <=> 2+b1+b1*q=0 <=> b1+b2=-2 \\
10+S_{4}=0 <=> 10+b1+b2+b3+b4=0<=> b1+b2+b3+b4=-10[/tex]
Deci:
b1+b2=-2 (1)
b1+b2+b3+b4=-10 (2)
Inlocuim prima relatie in cea de-a doua si obtinem o a treia relatie:
b3+b4=-10+2=-8 (3)
Luam relatiile (1) si (3) si dezvoltam dupa formula bn=b1*q^(n-1):
[tex]\left \{ {{b1+b2=-2 } \atop {b3+b4=-8 }} \right. <=> \left \{ {{b1+b1*q=-2} \atop {b1*q^2+b1*q^3=-8}} \right.[/tex]
[tex]<=> \left \{ {{b1(1+q)=-2} \atop {b1(q^2+q^3)=-8}} \right.[/tex]
Impartim membru cu membru:
[tex]\frac{b1(1+q)}{b1(q^2+q^3)} =\frac{-2}{-8} \\[/tex]
Simplificam si obtinem: [tex]\frac{b1(1+q)}{b1*q^2(1+q)} =\frac{-2}{-8} \\\frac{1}{q^2} = \frac{1}{4} => q=2[/tex]
Revenim la ecuatia : b1(1+q)=-2, unde q=2. Deci [tex]b1=\frac{-2}{1+q}=-\frac{2}{3}[/tex]
Deci S3=b1+b2+b3=b1+b1*q+b1*q^2=b1*(1+q+q^2)=[tex]-\frac{2}{3} *(1+2+4)=-\frac{2}{3} *7=-14/3[/tex]
