Question

У геометричній прогресії Б12=8 Б14=2 знайти Б10

Answer 1

$\displaystyle\bf\\\left \{ {{b_{14} =2} \atop {b_{12} =8}} \right. \\\\\\:\left \{ {{b_{1} \cdot q^{13} =2} \atop {b_{1} \cdot q^{11} =8}} \right. \\--------\\q^{2} =\frac{1}{4} \\\\q_{1} =-\frac{1}{2} \ \ \ ; \ \ \ q_{2} =\frac{1}{2} \\\\b_{12} =b_{10} \cdot q^{2} \\\\b_{10} =b_{12} : q^{2} \\\\1)\\\\b'_{10} =8:\Big(-\frac{1}{2} \Big)^{2}=8:\frac{1}{4} =8\cdot 4=32\\\\2)\\\\b''_{10} =8:\Big(\frac{1}{2} \Big)^{2}=8:\frac{1}{4} =8\cdot 4=32\\\\Otvet \ : \ b_{10} =32$

Answer 2

Ответ:

32  ................................

Объяснение:

$b_{14} = b_{12}$ · q² ⇒ q² = 0,25  ;  $b_{10} = b_{12}/q^{2} = 32$