Question

Exercitiile
IX
X
progresia geometrica

Answer 1

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[tex]\it b_1+b_2+b_3 = 21 \Rightarrow b_1+b_1q+b_1q^2=21 \Rightarrow \\\;\\ \Rightarrow b_1(1+q+q^2)=21 \ \ \ (1) \\\;\\ b_1b_2b_3=64 \Rightarrow b_1\cdot b_1q\cdot b_1q^2 =64 \Rightarrow b_1^3q^3=64 \Rightarrow (b_1q)^3=4^3 \Rightarrow \\\;\\ \Rightarrow b_1q=4 \Rightarrow b_1=\dfrac{4}{q}\ \ \ (2) \\\;\\ (1),\ (2) \Rightarrow \dfrac{4}{q}(1+q+q^2) =21|_{\cdot q} \Rightarrow 4(1+q+q^2)=21q\Rightarrow [/tex]

$\it \Rightarrow 4q^2-17q+4=0\Rightarrow q = 4$

Am reținut numai soluția pozitivă, deoarece progresia este crescătoare.

[tex]\it b_1= \dfrac{4}{q} =\dfrac{4}{4} =1 \\\;\\ b_2=b_1q=1\cdot 4 = 4 \\\;\\ b_3= b_1q^2=1\cdot4^2=16.[/tex]

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