Question

Fie progresia geometrica in care b1=3,b2=12. Aflati b4
Ajutor ! Va rog!

Answer 1

b2=b1*q

q=b2/b1

q=12/3

q=4

bn=b1*q^(n-1)

b4=b1*q^3

b4=3*4^3

b4=3*64

b4=192

Answer 2

Fie q rația progresiei geometrice.

$\it q=\dfrac{b_2}{b_1} =\dfrac{12}{3} =4$

$\it b_4=b_2\cdot q^2=12\cdot4^2=12\cdot16 =192$