Question

verifique se cada sequencia abaixo e pg e em caso afirmativo determine sua razao a) (5/2, 15/2, 75/16,...) b) (3/4, -1/2, 1/3,..) c) (1,4,9,16,..) d) (-72,24,-8,..) e) (5,5,5,..) f) (40,-20,-10,..)

Answer 1

Para que seja $\text{PG}$ devemos ter $\dfrac{a_2}{a_1}=\dfrac{a_3}{a_2}$

a) $\left(\dfrac{5}{2},\dfrac{15}{2},\dfrac{75}{16},\dots\right)$

$\dfrac{a_2}{a_1}=\dfrac{\dfrac{15}{2}}{\dfrac{5}{2}}=\dfrac{15}{2}\cdot\dfrac{2}{5}=\dfrac{30}{10}=3$

$\dfrac{a_3}{a_2}=\dfrac{\dfrac{75}{16}}{\dfrac{15}{2}}=\dfrac{75}{16}\cdot\dfrac{2}{15}=\dfrac{150}{240}=\dfrac{5}{8}$

Não é $\text{PG}$.

b) $\left(\dfrac{3}{4},-\dfrac{1}{2},\dfrac{1}{3},\dots\right)$

$\dfrac{a_2}{a_1}=\dfrac{\dfrac{-1}{2}}{\dfrac{3}{4}}=\dfrac{-1}{2}\cdot\dfrac{4}{3}=\dfrac{-4}{6}=\dfrac{-2}{3}$

$\dfrac{a_3}{a_2}=\dfrac{\dfrac{1}{3}}{\dfrac{-1}{2}}=\dfrac{1}{3}\cdot\dfrac{2}{-1}=\dfrac{-2}{3}$

É uma $\text{PG}$ de razão $q=\dfrac{-2}{3}$

c) $(1,4,9,16,\dots)$

$\dfrac{a_2}{a_1}=\dfrac{4}{1}=4$

$\dfrac{a_3}{a_2}=\dfrac{9}{4}$

Não é $\text{PG}$

d) $(-72,24,-8\dots)$

$\dfrac{a_2}{a_1}=\dfrac{24}{-72}=\dfrac{-1}{3}$

$\dfrac{a_3}{a_2}=\dfrac{-8}{24}=\dfrac{-1}{3}$

É uma $\text{PG}$ de razão $q=\dfrac{-1}{3}$

e) $(5,5,5,\dots)$

$\dfrac{a_2}{a_1}=\dfrac{5}{5}=1$

$\dfrac{a_3}{a_2}=\dfrac{5}{5}=1$

É uma $\text{PG}$ de razão $q=1$

f) $(40,-20,-10,\dots)$

$\dfrac{a_2}{a_1}=\dfrac{-20}{40}=\dfrac{-1}{2}$

$\dfrac{a_3}{a_2}=\dfrac{-10}{-20}=\dfrac{1}{2}$

Não é $\text{PG}$