Question

сумма в2+в3=30, в4-в2=90, найти в5​

Answer 1

$\displaystyle\bf\\\left \{ {{b_{4} -b_{2} =90} \atop {b_{2} + b_{3} =30}} \right. \\\\\\\left \{ {{b_{1}q^{3} -b_{1} q=90} \atop {b_{1} q+ b_{1} q^{2} =30}} \right.\\\\\\\left \{ {{b_{1} q\cdot(q^{2} -1)=90} \atop {b_{1} q\cdot(1+q)=30}} \right. \\\\\\:\left \{ {{b_{1} q\cdot(q -1)(q+1)=90} \atop {b_{1} q\cdot(1+q)=30}} \right. \\--------------\\q-1=3\\\\q=4$

$\displaystyle\bf\\b_{1} =\frac{30}{q\cdot(1+q)} =\frac{30}{4\cdot(1+4)} =1,5\\\\\\b_{5} =b_{1} \cdot q^{4} =1,5\cdot 4^{4} =1,5\cdot 256=384\\\\\\Otvet \ : \ b_{5} =384$