write a recursive rule for the sequence. NO LINKS!!!

Answer:
See below for answers and explanations
Step-by-step explanation:
Problem 55:
The sequence is geometric since each term is multiplied by 4, which is our common ratio. The recursive formula for a geometric sequence is [tex]a_n=a_1*r^{n-1}[/tex] where [tex]a_n[/tex] is the nth term of the sequence, [tex]a_1[/tex] is the first term of the sequence, and [tex]r[/tex] is the common ratio. Given [tex]r=4[/tex] and [tex]a_1=2[/tex], the recursive formula for the geometric sequence is [tex]a_n=2*4^{n-1}[/tex].
Problem 56:
Applying the same steps as the previous problem, we see that each term is getting multiplied by 1/2, so it's a geometric sequence with our common ratio as [tex]r=\frac{1}{2}[/tex] and the first term of the sequence as [tex]a_1=48[/tex]. Therefore, the recursive formula for the geometric sequence is [tex]a_n=48*\frac{1}{2}^{n-1}[/tex].
Problem 57:
Applying the same steps as the previous problem, we see that each term is getting multiplied by 1/3, so it's a geometric sequence with our common ratio as [tex]r=\frac{1}{3}[/tex] and the first term of the sequence as [tex]a_1=36[/tex]. Therefore, the recursive formula for the geometric sequence is [tex]a_n=36*\frac{1}{3}^{n-1}[/tex].
9514 1404 393
Answer:
55. a[1] = 2; a[n] = 4·a[n-1]
56. a[1] = 48; a[n] = (1/2)·a[n-1]
57. a[1] = 36; a[n] = (1/3)·a[n-1]
Step-by-step explanation:
A recursive rule comes in two parts: (1) the initial value; (2) the relation between a given value and the previous one.
For a geometric sequence, the form is pretty simple. The initial value is the first term. The present value is the last one multiplied by the common ratio.
a[1] = a_1
a[n] = r·a[n-1]
__
55. The common ratio is 8/2 = 4. The first term is 2. The recursive rule is ...
a[1] = 2; a[n] = 4·a[n-1]
__
56. The common ratio is 24/48 = 1/2. The first term is 48. The recursive rule is ...
a[1] = 48; a[n] = (1/2)·a[n-1]
__
57. The common ratio is 12/36 = 1/3. The first term is 36. The recursive rule is ...
a[1] = 36; a[n] = (1/3)·a[n-1]