Знайдіть суму п'яти перших членів геометричної прогресії якщо b4-b5=-12, b6-b4=18

Question 1

Question 2

$b_4-b_5=-12$
$b_6-b_4=18$
$S_5-$ ?

$b_n=b_1*q^{n-1}$
$b_4=b_1*q^3$
$b_5=b_1*q^4$
$b_6=b_1*q^5$

$\left \{ {{b_4-b_5=-12} \atop {b_6-b_4=18}} \right.$
$\left \{ {{b_1*q^3-b_1*q^4=-12} \atop {b_1*q^5-b_1*q^3=18}} \right.$
$\left \{ {{b_1*q^3(1-q)=-12} \atop {b_1*q^3(q^2-1)=18}} \right.$
$\left \{ {{-b_1*q^3(q-1)=-12} \atop {b_1*q^3(q-1)(q+1)=18}} \right.$
$\left \{ {{b_1*q^3(q-1)=12} \atop {12(q+1)=18}} \right.$
$\left \{ {{b_1*q^3(q-1)=12} \atop {2(q+1)=3}} \right.$
$\left \{ {{b_1*q^3(q-1)=12} \atop {q+1=1.5}} \right.$
$\left \{ {{b_1*q^3(q-1)=12} \atop {q=1.5-1}} \right.$
$\left \{ {{b_1*q^3(q-1)=12} \atop {q=0.5}} \right.$
$\left \{ {{q=0.5} \atop {b_1(q^3-q^4)=-12}} \right.$
$\left \{ {{q=0.5} \atop {b_1(0.125-0.0625)=-12}} \right.$
$\left \{ {{q=0.5} \atop {b_1*0.0625=-12}} \right.$
$\left \{ {{q=0.5} \atop {b_1=-192}} \right.$

$S_5= \frac{b_1*(1-q^5)}{1-q}$
$S_5= \frac{-192*(1- \frac{1}{32} )}{1- \frac{1}{2} } = \frac{-192* \frac{31}{32} }{ \frac{1}{2} } = \frac{-6*31}{0.5} = \frac{-186}{0.5} =-372$

Ответ: -372