Question

A sequência (x, 5, y) é uma progressão aritmética de razão r e a sequência (x, 4, y) é uma
progressão geométrica de razão q. Se x < y, então o quociente r/q é igual a:

Answer 1

$\large\boxed{\begin{array}{l}\rm PA(x,5,y)\\\rm r=5-x~~r=y-5\\\rm 5-x=y-5\\\rm x+y=5+5\\\rm x+y=10\\\rm PG(x,4,y)\\\rm q=\dfrac{4}{x}~~q=\dfrac{y}{4}\\\\\rm \dfrac{4}{x}=\dfrac{y}{4}\\\\\rm x\cdot y=4\cdot 4\\\rm x\cdot y=16\end{array}}$

$\large\boxed{\begin{array}{l}\sf Montando\,um\,sistema\,temos:\\\begin{cases}\rm x+y=10\\\rm xy=16\end{cases}\\\begin{cases}\rm y=10-x\\\rm xy=16\end{cases}\\\rm x\cdot(10-x)=16\\\rm 10x-x^2=16\\\rm x^2-10x+16=0\end{array}}$

$\large\boxed{\begin{array}{l}\rm \Delta=b^2-4ac\\\rm\Delta=(-10)^2-4\cdot1\cdot16\\\rm\Delta=100-64\\\rm\Delta=36\\\rm x=\dfrac{-b\pm\sqrt{\Delta}}{2a}\\\\\rm x=\dfrac{-(-10)\pm\sqrt{36}}{2\cdot1}\\\\\rm x=\dfrac{10\pm6}{2}\begin{cases}\rm x_1=\dfrac{10+6}{2}=\dfrac{16}{2}=8\\\\\rm x_2=\dfrac{10-6}{2}=\dfrac{4}{2}=2\end{cases}\end{array}}$

$\large\boxed{\begin{array}{l}\rm para\,x=8\\\rm y=10-x\\\rm y=10-8\\\rm y=2\\\rm para\,x=2\\\rm y=10-2\\\rm y=8\end{array}}$

$\large\boxed{\begin{array}{l}\rm Como\,x<y~ent\tilde ao~x=2~e~y=8\end{array}}$

$\large\boxed{\begin{array}{l}\rm para\,PA: (2,5,8)\longrightarrow r=5-2=3\\\rm para\,PG:(2,4,8)\longrightarrow q=\dfrac{4}{2}=2\\\rm \dfrac{r}{q}=\dfrac{3}{2}\\\huge\boxed{\boxed{\boxed{\boxed{\rm\dagger\red{\maltese}~\blue{alternativa~d}}}}}\end{array}}$