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Write a recursive rule for the sequence. NO LINKS!!!​

Write a recursive rule for the sequence. NO LINKS!!!​-example-1
User Karcsi
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2 Answers

2 votes

Answer:

See below for answers and explanations

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
a_n=a_1*r^(n-1) where
a_n is the nth term of the sequence,
a_1 is the first term of the sequence, and
r is the common ratio. Given
r=4 and
a_1=2, the recursive formula for the geometric sequence is
a_n=2*4^(n-1).

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
r=(1)/(2) and the first term of the sequence as
a_1=48. Therefore, the recursive formula for the geometric sequence is
a_n=48*(1)/(2)^(n-1).

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
r=(1)/(3) and the first term of the sequence as
a_1=36. Therefore, the recursive formula for the geometric sequence is
a_n=36*(1)/(3)^(n-1).

User Jdennison
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1 vote

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]

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]

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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]

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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]

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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]

User Zheek
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