Question

Calcule o r das seguintes P.A abaixo:
a) ( 2,5,8,11,14...)
b) ( 3,4,5,6,7...)
c) ( 6,6,6,6,6,...
d) ( 10,8,6,4,2,0...)

Answer 1

a) r = 5 - 2 = 3
b) r = 4 - 3 = 1
c) r = 6 - 6 = 0
d) r = 8 - 10 = -2

Answer 2

Formula da P.A.: 
$\boxed{\begin{array}{c}\mathsf{a_{n}=a_{1}+(n-1)\cdot r} \end{array}} \\ \\ a_{n}\rightarrow ultimo~termo~da~alternativa\\ a_{1}\rightarrow 1^{\circ} \\ n\rightarrow posi\c{C}\~ao~do~termo~da~p.a. \\ r\rightarrow raz\~ao$


Com base nesta informação vamos descobrir o r (razão da P.A.) das alternativas, assim:

a) $( 2,5,8,11,14...) \\ \\ 14=2+(5-1)\cdot r \\ 14=2+4\cdot r \\ 14-2=4\cdot r \\ 12=4\cdot r \\ \frac{12}{4}=r \\ \\ \boxed{\begin{array}{c}\mathsf{r=3} \end{array}}$

b) $( 3,4,5,6,7...) \\ \\ 7=3+(5-1)\cdot r \\ 7-3=(5-1)\cdot r \\ 4=4\cdot r \\ \frac{4}{4}=r \\ \\ \boxed{\begin{array}{c}\mathsf{r=1} \end{array}}$

c) $( 6,6,6,6,6,...) \\ \\ 6=6+(5-1)\cdot r \\ 6-6=(5-1)\cdot r \\ 0=4\cdot r \\ \frac{0}{4} =r \\ \\ \boxed{\begin{array}{c}\mathsf{r=0} \end{array}}$

d) $( 10,8,6,4,2,0...) \\ \\ 0=10+(6-1)\cdot r \\ 0-10=(6-1)\cdot r \\ -10=5\cdot r \\ \frac{-10}{5} = r \\ \\ \boxed{\begin{array}{c}\mathsf{r=-2} \end{array}}$

Espero ter ajudado, qualquer dúvida comente embaixo! :)