Question

Знайдіть перший член та знаменник геометричної прогресії (bn), якщо b2 b4 = 36 i b3 + b5 = 8.

Answer 1

Объяснение:

$\left \{ {{b_2*b_4=36} \atop {b_3+b_5=8}} \right.\ \ \ \ \ \left \{ {{b_1q*b_1q^3=36} \atop {b_1q^2+b_1q^4=8}} \right. \ \ \ \ \ \left \{ {{b_1^2q^4=36} \atop {b_1q^2+b_1q^4=8}}\right.\ \ \ \ \ \left \{ {{(b_1q^2)^2=6^2} \atop {b_1q^2*(1+q^2)=8}} \right.\ \ \ \ \ \left \{ {{b_1q^2=б6} \atop {b_1q^2*(1+q^2)=8}} \right. .$

$1)\ \left \{ {{b_1q^2=6} \atop {6*(1+q^2)=8\ |:6}} \right. \ \ \ \ \ \left \{ {{b_1q^2=6} \atop {1+q^2=\frac{4}{3} }} \right. \ \ \ \ \ \left \{ {{b_1q^2=6} \atop {q^2=\frac{1}{3} }} \right.\ \ \ \ \left \{ {b_*\frac{1}3}=6\ |*3 } \atop {q=б\frac{\sqrt{3} }{3} }} \right. \ \ \ \ \ \left \{ {{b_1=18} \atop {q=б\frac{\sqrt{3} }{3}}} \right. .$

$2)\ \left \{ {{b_1q^2=-6} \atop {-6*(1+q^2)=8\ |:6}} \right. \ \ \ \ \ \left \{ {{b_1q^2=6} \atop {1+q^2=-\frac{4}{3} }} \right. \ \ \ \ \ \left \{ {{b_1q^2=6} \atop {q^2=-\frac{7}{3}\notin }} \right..$

$OTBET:\ (b_1;q)=(18;\frac{\sqrt{3} }{3} )=(18;-\frac{\sqrt{3} }{3}).$