An arithmetic progression is a sequence that has a common difference between consecutive terms.
Question 6:
The recursive function is given as:
[tex]f(0) = 19\\f(n) = f(n - 1) -6;[/tex] for [tex]n \ge 1[/tex]
The definition of nth term is given as:
[tex]f(n) = 19 - 6n[/tex] for [tex]n > 0[/tex]
A proof that the functions represent the same sequence is as follows:
Set [tex]n=1[/tex] in
[tex]f(0) = 19\\f(n) = f(n - 1) -6;[/tex]
[tex]f(1) = f(1 -1) - 6 = f(0) - 6[/tex]
[tex]f(1) = 19 - 6 = 13[/tex]
Set [tex]n = 2[/tex]
[tex]f(2) = f(2 -1) - 6 = f(1) - 6[/tex]
[tex]f(2) = 13 - 6 = 7[/tex]
Set [tex]n = 3[/tex]
[tex]f(3) = f(3 -1) - 6 = f(2) - 6[/tex]
[tex]f(3) = 7 - 6 = 1[/tex]
Write out the equations
[tex]f(1) = 19 - 6 = 13[/tex]
[tex]f(2) = 13 - 6 = 7[/tex]
[tex]f(3) = 7 - 6 = 1[/tex]
Rewrite as:
[tex]f(1) = 19 - 6 = 13[/tex]
[tex]f(2) = 19 - 6 - 6 = 7[/tex]
[tex]f(3) = 19 - 6 - 6 -6= 1[/tex]
Rewrite again as:
[tex]f(1) = 19 -6\times 1 = 13[/tex]
[tex]f(2) = 19 -6\times 2 = 7[/tex]
[tex]f(3) = 19 -6\times 3 = 1[/tex]
Replace the numbers in brackets with n
[tex]f(n) = 19 -6\times n[/tex]
[tex]f(n) = 19 -6n[/tex]
Hence, it is proved that
[tex]f(n) = 19 -6n[/tex] and
[tex]f(0) = 19\\f(n) = f(n - 1) -6;[/tex] represents the same sequence
The definition I will choose for f(20) is:
[tex]f(20) = 19 - 6 \times 20[/tex]
This is gotten from [tex]f(n) = 19 -6n[/tex].
I choose the definition because it is faster to calculate f(20) directly, than using recursion
Question 7:
Given that:
[tex]T_1 = 20[/tex]
[tex]T_2 = 16[/tex]
First, we calculate the common difference
[tex]d = T_2 - T_1[/tex]
[tex]d = 16 - 20 =-4[/tex]
The 500th term is then calculated using the nth term of an arithmetic progression formula.
[tex]T_n = T_1 + (n - 1) \times d[/tex]
So, we have:
[tex]T_{500} = 20+ (500 - 1) \times -4[/tex]
[tex]T_{500} = 20+ 499 \times -4[/tex]
[tex]T_{500} = 20 -1996[/tex]
[tex]T_{500} = -1976[/tex]
Hence, the 500th term is -1976.
Read more about arithmetic progression at:
https://brainly.com/question/22445557
