Question

1-Encontre a razão das PG:
a)(√2,2,..)
b)(xy,xy^3,...)

2- Uma PG de cindo termos, na qual a1= x/y^3 e q= y^2

Answer 1

$\Large\boxed{\begin{array}{l}\rm a)~\sf (\sqrt{2},2\dotsc)\\\sf q=\dfrac{2}{\sqrt{2}}\\\\\sf q=\dfrac{2}{\sqrt{2}}\cdot\dfrac{\sqrt{2}}{\sqrt{2}}\\\\\sf q=\dfrac{2\sqrt{2}}{\sqrt{2^2}}\\\\\sf q=\dfrac{\diagup\!\!\!\!2\sqrt{2}}{\diagup\!\!\!\!2}\\\\\sf q=\sqrt{2}\end{array}}$

$\Large\boxed{\begin{array}{l}\rm b)~\sf(xy,xy^2\dotsc)\\\sf q=\dfrac{\backslash\!\!\!x\diagup\!\!\!\!y^2}{\backslash\!\!\!x\diagup\!\!\!y}\\\\\sf q=y\end{array}}$

$\Large\boxed{\begin{array}{l}\rm 2)~\sf a_1=\dfrac{x}{y^3}~~q=y^2\\\\\sf a_2=a_1\cdot q\\\sf a_2=\dfrac{x}{\diagup\!\!\!\!y^3}\cdot \diagup\!\!\!\!\!y^2\\\\\sf a_2=\dfrac{x}{y}\\\\\sf a_3=a_2\cdot q\\\sf a_3=\dfrac{x}{\diagup\!\!\!\!y}\cdot \diagup\!\!\!\!y^2\\\\\sf a_3=xy\\\sf a_4=a_3\cdot q\\\sf a_4=xy\cdot y^2=xy^3\\\sf a_5=a_4\cdot q\\\sf a_5= xy^3\cdot y^2\\\sf a_5=xy^5\end{array}}$

$\Large\boxed{\begin{array}{l}\sf\bigg(\dfrac{x}{y^3},\dfrac{x}{y},xy,xy^3,xy^5\bigg)\end{array}}$