Question

геометриялық прогрессияның b1+b2=16, b2+b3=36 прогрессия ның алғашқы үш мүшесін табыныз​

Answer 1

$b_{1}+b_{2}=16\\\\b_{1}+b_{1}q=16\\\\b_{1}(1+q)=16\\\\\\b_{2}+b_{3}=36\\\\b_{1}q+b_{1}q^{2}=36\\\\b_{1}q(1+q)=36\\\\\\:\left \{ {{b_{1}q(1+q)=36 } \atop {b_{1}(1+q)=16 }} \right. \\q=2,25\\\\b_{1} =\frac{16}{1+q}=\frac{16}{1+2,25}=\frac{16}{3,25} =16:3\frac{1}{4}=16:\frac{13}{4}=16*\frac{4}{13}=\frac{64}{13}=4\frac{12}{13}$

$b_{2}=b_{1}*q=4\frac{12}{13}*2,25=\frac{64}{13}*\frac{9}{4}=\frac{16*9}{13} =\frac{144}{13}=11\frac{1}{13} \\\\b_{3}=36-11\frac{1}{13}=24\frac{12}{13}\\\\Otvet:\boxed{4\frac{12}{13};11\frac{1}{13};24\frac{12}{13}}$