Question

determine a razão de cada pg a) (2√5, -10, 10√5...) b) (-1, -√7, -7,...) c) (10^-1, 10)

Answer 1

A razão (q) de uma PG é dada pelo quociente entre um termo e seu antecessor.

$\boxed{q~=~\dfrac{a_{n}}{a_{n-1}}}$

a)

$q~=~\dfrac{a_3}{a_2}\\\\\\q~=~\dfrac{10\sqrt{5}}{-10}\\\\\\q~=~\dfrac{1\sqrt{5}}{-1}\\\\\\\boxed{q~=\,-\sqrt{5}}$

b)

$q~=~\dfrac{a_2}{a_1}\\\\\\q~=~\dfrac{-\sqrt{7}}{-1}\\\\\\\boxed{q~=~\sqrt{5}}$

c)

$q~=~\dfrac{a_2}{a_1}\\\\\\q~=~\dfrac{10}{10^{-1}}\\\\\\q~=~10^{1-(-1)}\\\\\\q~=~10^{1+1}\\\\\\q~=~10^2\\\\\\\boxed{q~=~100}$