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set topic: finding arithmetic and geometric means and making meaning of…

Question

set
topic: finding arithmetic and geometric means and making meaning of rational exponents
you may have found arithmetic and geometric means in your prior work. finding arithmetic and geometric means requires finding values of a sequence between given values from non - consecutive terms. in each of the sequences below determine the means and show how you found them.
find the arithmetic means for the following. show your work.
10.

x123
y511

11.

x12345
y18- 310

12.

x1234567
y12- 6

find the geometric means for the following. show your work.
13.

x123
y312

14.

x1234
y7875

15.

x123456
y4972

fill in the tables of values and find the factor used to move between whole number values, f_w, as well as the factor, f_c, used to move between each column of the table.
16.

x01/213/22
y416

d. f_w =
e. f_c =
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Explanation:

Step1: Recall arithmetic - mean formula for an arithmetic sequence

For an arithmetic sequence \(a_n=a_1+(n - 1)d\), where \(a_n\) is the \(n\)th term, \(a_1\) is the first - term, \(n\) is the number of terms, and \(d\) is the common difference.

Step2: Solve problem 10

We have \(a_1 = 5\), \(a_4=11\), and \(n = 4\). Using the formula \(a_n=a_1+(n - 1)d\), we substitute the values: \(11 = 5+(4 - 1)d\). Then \(11-5 = 3d\), so \(6 = 3d\) and \(d = 2\). The arithmetic means are \(a_2=a_1 + d=5 + 2=7\) and \(a_3=a_2 + d=7 + 2 = 9\).

Step3: Solve problem 11

We know \(a_1 = 18\), \(a_5 = 10\), \(n = 5\). Using \(a_n=a_1+(n - 1)d\), we get \(10=18+(5 - 1)d\). Then \(10 - 18=4d\), so \(- 8 = 4d\) and \(d=-2\). \(a_2=a_1 + d=18-2 = 16\), \(a_3=a_2 + d=16-2 = 14\), \(a_4=a_3 + d=14-2 = 12\).

Step4: Solve problem 12

We have \(a_1 = 12\), \(a_7=-6\), \(n = 7\). Using \(a_n=a_1+(n - 1)d\), we have \(-6=12+(7 - 1)d\). Then \(-6 - 12=6d\), so \(-18 = 6d\) and \(d=-3\). \(a_2=a_1 + d=12-3 = 9\), \(a_3=a_2 + d=9-3 = 6\), \(a_4=a_3 + d=6-3 = 3\), \(a_5=a_4 + d=3-3 = 0\), \(a_6=a_5 + d=0-3=-3\).

Step5: Recall geometric - mean formula for a geometric sequence

For a geometric sequence \(a_n=a_1r^{n - 1}\), where \(a_n\) is the \(n\)th term, \(a_1\) is the first - term, \(n\) is the number of terms, and \(r\) is the common ratio.

Step6: Solve problem 13

We have \(a_1 = 3\), \(a_3 = 12\). Using \(a_n=a_1r^{n - 1}\), for \(n = 3\), \(a_3=a_1r^{2}\), so \(12 = 3r^{2}\). Then \(r^{2}=4\), and \(r = 2\) (we consider the positive value for a common - ratio in a simple geometric sequence context). The geometric mean \(a_2=a_1r=3\times2 = 6\).

Step7: Solve problem 14

We have \(a_1 = 7\), \(a_4 = 875\). Using \(a_n=a_1r^{n - 1}\), for \(n = 4\), \(a_4=a_1r^{3}\), so \(875 = 7r^{3}\). Then \(r^{3}=\frac{875}{7}=125\), and \(r = 5\). \(a_2=a_1r=7\times5 = 35\), \(a_3=a_2r=35\times5 = 175\).

Step8: Solve problem 15

We have \(a_1 = 4\), \(a_6 = 972\). Using \(a_n=a_1r^{n - 1}\), for \(n = 6\), \(a_6=a_1r^{5}\), so \(972 = 4r^{5}\). Then \(r^{5}=\frac{972}{4}=243\), and \(r = 3\). \(a_2=a_1r=4\times3 = 12\), \(a_3=a_2r=12\times3 = 36\), \(a_4=a_3r=36\times3 = 108\), \(a_5=a_4r=108\times3 = 324\).

Step9: Solve problem 16

We know when \(x = 0\), \(y = 4\) and when \(x = 1\), \(y = 16\). For a geometric - like relationship \(y=a\cdot b^x\), when \(x = 0\), \(y=a\), so \(a = 4\). When \(x = 1\), \(y=ab\), and since \(a = 4\) and \(y = 16\), then \(4b = 16\), \(b = 4\).
When \(x=\frac{1}{2}\), \(y=4\times4^{\frac{1}{2}}=4\times2 = 8\). When \(x=\frac{3}{2}\), \(y=4\times4^{\frac{3}{2}}=4\times8 = 32\).
The factor \(F_w\) (between whole - number values of \(x\)) is \(4\) (since when \(x\) changes from \(0\) to \(1\), \(y\) changes from \(4\) to \(16\)). The factor \(F_c\) (between each column) is also \(4\).

Answer:

  1. Arithmetic means: 7, 9
  2. Arithmetic means: 16, 14, 12
  3. Arithmetic means: 9, 6, 3, 0, - 3
  4. Geometric mean: 6
  5. Geometric means: 35, 175
  6. Geometric means: 12, 36, 108, 324
  7. \(y\) values: 8, 32; \(F_w = 4\), \(F_c = 4\)