[tex]\displaystyle\bf\\b_{1}=-6\\\\b_{4}=0,75\\\\b_{2}=? \ \ \ ; \ \ \ b_{3}=?\\\\\\b_{4}=b_{1}\cdot q^{3}\\\\\\q^{3} =\frac{b_{4} }{b_{1} } =\frac{0,75}{-6} =-0,125\\\\\\q=\sqrt[3]{-0,125} =-\sqrt[3]{0,5^{3} } =-0,5\\\\\\b_{2} =b_{1} \cdot q=-6\cdot (-0,5)=3\\\\\\b_{3} =b_{2} \cdot q=3\cdot (-0,5)=-1,5\\\\\\Otvet \ : \ -6 \ \ ; \ \ 3 \ \ ; \ \ -1,5 \ \ ; \ \ 0,75[/tex]