Question

Fie bn o progresie geometrica cu:
a) b3=2 si b5=4
b) {b1+ b2=3 , b3 +b4=12
Determinanți b1 si q

Answer 1

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