3. The sequence 6, 18, 54, 162, … shows the number of pushups Kendall did each week, starting with her first week of exercising. (a) What is the recursive rule for the sequence? (b) What is the iterative rule for the sequence?

Answer:

Please check the explanation.

Step-by-step explanation:

(a) What is the recursive rule for the sequence?

Given the sequence

6, 18, 54, 162,...

Please note that each successive term is 3 times the previous term.

i.e. 18 = 6×3, 54 = 18×3, 162 = 54×3

Here, r = 3 is the common multiplier

In other words, every next term can be obtained by multiplying the previous term by a number 'r', using the recursive formula

[tex]a_n=a_{n-1}\times r[/tex]

Here, aₙ is r times the previous term aₙ₋1.

For example, in order to determine the 2nd term a₂, just put n=2 in the recursive formula

[tex]a_n=a_{n-1}\times r[/tex]

so substituting r = 3, and n = 1 in the recursive formula

[tex]a_2=a_{2-1}\times 3[/tex]

[tex]a_2=a_1\times \:3\:\:\:\:\:\:[/tex]

as

a₁ = 6

so

[tex]a_2=6\times \:3\:\:\:\:\:\:[/tex]

[tex]a_2=18[/tex]

(b) What is the iterative rule for the sequence?

Given the sequence

6, 18, 54, 162,...

A geometric sequence has a constant ratio 'r' and is defined as

[tex]a_n=a_1\cdot r^{n-1}[/tex]

[tex]\frac{18}{6}=3,\:\quad \frac{54}{18}=3,\:\quad \frac{162}{54}=3[/tex]

The ratio of all the adjacent terms is the same and equal to, so

r = 3

as

a₁ = 6

so substituting r = 3 and a₁ = 6 in the nth term of the sequence

[tex]a_n=a_1\cdot r^{n-1}[/tex]

[tex]a_n=6\cdot \:3^{n-1}[/tex]

Thus, the iterative rule for the sequence is

a) The recursive rule for the sequence is [tex]a_n=3a_{n-1}[/tex]

b) The iterative rule for the sequence is [tex]a_n=2(3^n)[/tex]

The given sequence is:

6, 18, 54, 162, …

The common ratio, r = 18/6

r  =  3

Let the last term be [tex]a_n[/tex]

Let the preceding term be [tex]a_{n-1}[/tex]

The recursive rule for the sequence is given as:

[tex]a_n=a_{n-1}r\\\\a_n=3a_{n-1}[/tex]

b) The iterative rule is gotten by using the formula:

[tex]a_n=ar^{n-1}[/tex]

Substitute a = 6 and r = 3 into the formula above

[tex]a_n=6(3)^{n-1}\\\\a_n=\frac{6}{3}(3^n )\\\\a_n=2(3^n)[/tex]

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