Question

Progresia geometrica (bn)n≥1 este definita de b4 = 8 si
S2/S4=1/5
Calculati q, b1.

Answer 1

Pentru a evita cazurile irelevante, considerăm rația  q ∈ ℝ*,  q ≠ ± 1.

[tex]\it \ b_4=8 \Rightarrow b_3=\dfrac{8}{q},\ \ b_2=\dfrac{8}{q^2},\1 \ b_1=\dfrac{8}{q^3} \\ \\ \\ \dfrac{S_2}{S_4} = \dfrac{1}{5} \Rightarrow 5S_2=S_4 \Rightarrow 5S_2=S_2+b_3+b_4\Rightarrow 4S_2= b_3+b_4 \Rightarrow \\ \\ \\ \Rightarrow 4(b_1+b_2) = b_3+b_4 \Rightarrow 4\left(\dfrac{8}{q^3}+\dfrac{8}{q^2}\right) = \dfrac{8}{q} +8 \Rightarrow [/tex]

[tex]\it 4\cdot8 \left(\dfrac{1}{q^3}+\dfrac{1}{q^2}\right) = 8\cdot\left(\dfrac{1}{q} +1\right)|_{:8} \Rightarrow 4\cdot\dfrac{1+q}{q^3} =\dfrac{1+q}{q} |_{\cdot \frac{q}{1+q}} \Rightarrow \\ \\ \\ \Rightarrow \dfrac{4}{q^2} =1 \Rightarrow q^2=4 \Rightarrow q= \pm2 \\ \\ \\ b_1= \dfrac{8}{q^3} = \dfrac{8}{(\pm2)^3} = \dfrac{8}{\pm8} = \pm1 [/tex]

Avem două soluții:

[tex]\it b_1= -1,\ \ q = -2 \\ \\ \\ b_1 = 1,\ \ q = 2[/tex]