Question

Dados os termos a1 (primeiro termo) e a razão q determine os cinco primeiros termos de cada P.G
a) a1= 4 e q=3
b) a1=20 e q= -2
c) a1=10 e q= 1/5​

Answer 1

Na PG, todo novo termo ($a_n$) é dado pelo produto entre a razão (q) e seu termo anterior ($a_{n-1}$).

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

Assim, por exemplo, o 3º termo será calculado multiplicando-se o 2º termo pela razão da PG.

Dito isso, podemos calcular o que é pedido.

a)

$a_2~=~a_1\cdot q~=~4\cdot 3~~ \Rightarrow~~\boxed{a_2~=~12}\\\\\\a_3~=~a_2\cdot q~=~12\cdot 3~~ \Rightarrow~~\boxed{a_3~=~36}\\\\\\a_4~=~a_3\cdot q~=~36\cdot 3~~ \Rightarrow~~\boxed{a_4~=~108}\\\\\\a_5~=~a_4\cdot q~=~108\cdot 3~~ \Rightarrow~~\boxed{a_5~=~324}\\\\\\PG:~~ \{4~,~12~,~36~,~108~,~324\}$

b)

$a_2~=~a_1\cdot q~=~20\cdot (-2)~~ \Rightarrow~~\boxed{a_2~=\,-40}\\\\\\a_3~=~a_2\cdot q~=\,-40\cdot (-2)~~ \Rightarrow~~\boxed{a_3~=~80}\\\\\\a_4~=~a_3\cdot q~=~80\cdot (-2)~~ \Rightarrow~~\boxed{a_4~=\,-160}\\\\\\a_5~=~a_4\cdot q~=\,-160\cdot (-2)~~ \Rightarrow~~\boxed{a_5~=~320}\\\\\\PG:~\{20~,\,-40~,~80~,\,-160~,~320\}$

c)

$a_2~=~a_1\cdot q~=~10\cdot \dfrac{1}{5}~=~\dfrac{10}{5}~~ \Rightarrow~~\boxed{a_2~=~2}\\\\\\a_3~=~a_2\cdot q~=~2\cdot \dfrac{1}{5}~~ \Rightarrow~~\boxed{a_3~=~\dfrac{2}{5}}~~ ou~~ \boxed{a_3~=~0,4}\\\\\\a_4~=~a_3\cdot q~=~\dfrac{2}{5}\cdot \dfrac{1}{5}~~ \Rightarrow~~\boxed{a_4~=~\dfrac{2}{25}}~~ ou~~ \boxed{a_4~=~0,08}\\\\\\a_5~=~a_4\cdot q~=~\dfrac{2}{25}\cdot \dfrac{1}{5}~~ \Rightarrow~~\boxed{a_5~=~\dfrac{2}{125}}~~ ou~~ \boxed{a_5~=~0,016}$

$PG:~\left\{10~,~2~,~\dfrac{2}{5}~,~\dfrac{2}{25}~,~\dfrac{2}{125}\right\}~~ou~~ \{10~,~2~,~0,4~,~0,08~,~0,016\}$

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