Într-o progresie geometrică, bn = b1 × q^(n-1)
În cazul nostru:
a) b3 = b1 × q^2 = 2
b5 = b1 × q^4 = 4
b5/b3 = b1 × q^4/(b1 × q^2) = q^2 = 4/2 = 2
Atunci q = radical 2
Dar b1 × q^2 = b1 × 2 = 2 => b1 = 1
R: b1 = 1, q= radical 2
b) b1 + b2 = 3 => b1 + b1×q = b1(1 + q) = 3
b3 + b4 = b1×q^2 + b1×q^3 = b1×q^2×(1 +q) = 12
(b3 + b4)/(b1 + b2) = b1×q^2×(1+q)/b1×(1+q) = q^2 = 12/3 = 4
Atunci q = radical 4 = 2
Dar b1(1+q) = 3 => b1(1+2) = 3 => 3b1 = 3 => b1 = 1
R: b1 = 1, q = 2
