Odpowiedź:
[tex]\huge\boxed{m = -1 \ \vee \ m = 16}[/tex]
Szczegółowe wyjaśnienie:
Z własności ciągu geometrycznego:
[tex]a_2^{2} = a_1\cdot a_3\\\\a_1 = m, \ a_2 = -4, \ a_3=m-15\\\\\\(-4)^{2} = m(m-15)\\\\16 = m^{2}-15 \ m\\\\m^{2}-15m-16 = 0\\\\\Delta = b^{2}-4ac = (-15)^{2}-4\cdot1\cdot16 = 225+64 = 289\\\\\sqrt{\Delta} = \sqrt{289} = 17\\\\m_1 = \frac{-b-\sqrt{\Delta}}{2a} = \frac{15-17}{2\cdot1} = \frac{-2}{2} = \boxed{-1}\\\\m_2 = \frac{-b+\sqrt{\Delta}}{2a} = \frac{15+17}{2} = \frac{32}{2} = \boxed{16}[/tex]
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Odpowiedź:
[tex]\huge\boxed{m = -1 \ \vee \ m = 16}[/tex]
Szczegółowe wyjaśnienie:
Z własności ciągu geometrycznego:
[tex]a_2^{2} = a_1\cdot a_3\\\\a_1 = m, \ a_2 = -4, \ a_3=m-15\\\\\\(-4)^{2} = m(m-15)\\\\16 = m^{2}-15 \ m\\\\m^{2}-15m-16 = 0\\\\\Delta = b^{2}-4ac = (-15)^{2}-4\cdot1\cdot16 = 225+64 = 289\\\\\sqrt{\Delta} = \sqrt{289} = 17\\\\m_1 = \frac{-b-\sqrt{\Delta}}{2a} = \frac{15-17}{2\cdot1} = \frac{-2}{2} = \boxed{-1}\\\\m_2 = \frac{-b+\sqrt{\Delta}}{2a} = \frac{15+17}{2} = \frac{32}{2} = \boxed{16}[/tex]