Question

ratia unei progresii geometrice daca am b4=27

Answer 1

$\dfrac{..}{..}\ \ b_4=27,\ b_7 = 729$

$b_7=b_4\cdot q^3 \Longrightarrow q^3=\dfrac{b_7}{b_4} = \dfrac{729}{27} =27$

$q^3=27 \Longrightarrow q^3=3^3 \Longrightarrow q=3$

$b_1=\dfrac{b_4}{q^3} =\dfrac{27}{27} = 1$