Question

Progresie geometrică…

Answer 1

Răspuns:

$b1=1$

$b2+b3=12$

$=b1*q+b1*q^2=12$

$q^2+q-12=0$

$q^2+4q-3q-12=0$

$q(q+4)-3(q+4)$

$(q+4)(q-3)=0=>q1=-4;q2=3$

Dar b2+b3>0 => q = 3

Answer 2

$\it b_2+b_3=12 \Rightarrow b_2+b_2q=12 \Rightarrow b_2(1+q)=12\ \ \ \ \ \ \ (1)\\ \\ b_2=b_1q \Rightarrow b_2=1\cdot q \Rightarrow b_2=q\ \ \ \ \ \ \ (2)\\ \\ (1),\ (2) \Rightarrow q(1+q)=12 \Rightarrow q(q+1)=12 = -4\cdot(-3)=3\cdot4 \Rightarrow q\in\{-4,\ 3\}$