• Matéria: Matemática
  • Autor: merd87
  • Perguntado 3 anos atrás

Qual é a razão da PG (x, x + 4, x + 6, ...)?​

Respostas

respondido por: CyberKirito
0

\large\boxed{\begin{array}{l}\rm (x,x+4,x+6\dotsc)\\\rm a_2^2=a_1\cdot a_3\\\rm (x+4)^2=x\cdot(x+6)\\\rm x^2+8x+16=x^2+6x\\\rm \diagup\!\!\!\!\!x^2-\diagup\!\!\!\!x^2+8x-6x=-16\\\rm 2x=-16\\\rm x=-\dfrac{16}{2}\\\\\rm x=-8\\\rm  PG~(-8,-4,-2)\\\rm q=\dfrac{a_2}{a_1}\\\\\rm q=\dfrac{-4\div(-4)}{-8\div(-4)}\\\\\rm q=\dfrac{1}{2}\end{array}}

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