Answer:
a) 3, 6, 9, 12, 15,...,[tex]3\cdot n[/tex], b) 4, 7, 10, 13, 16,...,[tex]3\cdot n +1[/tex], c) Both sequences are arithmetic.
Step-by-step explanation:
a) The sequence of natural numbers which are multiplied by 3 are represented by the function [tex]f(n) = 3\cdot n[/tex], [tex]n\in \mathbb{N}[/tex]. Let see the first five elements of the sequence: 3, 6, 9, 12, 15,...
b) The sequence of natural numbers which are multiplied by 3 and added to 1 is represented by the function [tex]f(n) = 3\cdot n + 1[/tex], [tex]n\in \mathbb{N}[/tex]. Let see the first five elements of the sequence: 4, 7, 10, 13, 16,...
c) Both sequences since differences between consecutive elements is constant. Let prove this statement:
(i) [tex]f(n) = 3\cdot n[/tex]
[tex]\Delta f = f(n+1) -f(n)[/tex]
[tex]\Delta f = 3\cdot (n+1) -3\cdot n[/tex]
[tex]\Delta f = 3[/tex]
(ii) [tex]f(n) = 3\cdot n +1[/tex]
[tex]\Delta f = f(n+1)-f(n)[/tex]
[tex]\Delta f = [3\cdot (n+1)+1]-(3\cdot n+1)[/tex]
[tex]\Delta f = 3[/tex]
Both sequences are arithmetic.
