Question

В геометрической прогрессии b1 + b2 = 15, b2 + b3 = 30. Найдите первые пять членов прогрессии сорчнооо помагите

Answer 1

Answer 2

$\left \{ {{b_{1}+b_{2}=15} \atop {b_{2}+b_{3}=30}} \right.\\\\\left \{ {{b_{1} +b_{1}q=15} \atop {b_{1}q+b_{1}q^{2}=30}} \right.\\\\:\left \{ {{b_{1}q(1+q)=30 } \atop {b_{1}(1+q)=15 }} \right.\\-------\\q=2\\\\b_{1}=15:(1+q)=15:(1+2)=5\\\\b_{2}=b_{1}*q=5*2=10\\\\b_{3}=b_{2}*q=10*2=20\\\\b_{4} =b_{3}*q=20*2=40\\\\b_{5}=b_{4} *q=40*2=80\\\\Otvet:\boxed{5;10;20;40;80...}$