Question

Va rogggg cat de repede daca ma puteti ajuta!!!

Multumesc!!
Determinati trei numere in progresie geometrica cu sumaegala cu 19 si raportul intre primul si al treilea termen 1/4.

Answer 1

Am atasat o rezolvare.

Answer 2

$b_1+b_2+b_3 = 19\\ \\\dfrac{b1}{b3} = \dfrac{1}{4} \Rightarrow \dfrac{b_1}{b_1\cdot q^2} = \dfrac{1}{4} \Rightarrow \dfrac{1}{q^2} = \dfrac{1}{4} \Rightarrow \text{alegem }q = 2\\ \\ b_1+b_1\cdot q+b_1\cdot q^2 = 19 \\ b_1+2b_1+4b_1 = 19 \\ 7b_1 = 19\\ b_1 = \dfrac{19}{7} \\ \\ \Rightarrow b_1 = \dfrac{19}{7},~~b_2 = \dfrac{38}{7},\quad b_3 = \dfrac{76}{7}$