Find the three arithmetic means in this sequence

-5.6, -0.2, 5.2, 10.6

Step-by-step explanation:

This is a geometric sequence so use its general equation to solve for the missing values.

     aₙ = a₁ + d(n - 1)

aₙ is the nth after the first number you're trying to find.

a₁ is the first number.

d is your scaling value. (n-1) is just there to help you calculate the previous values.

You have the first and last number (the sixth number which is 16). Use those to find your scaling value then it's just plug and chugg after that.

-5.6, -0.2, 5.2, 10.6

Step-by-step explanation:

This is a geometric sequence so use its general equation to solve for the missing values.

     aₙ = a₁ + d(n - 1)

aₙ is the nth after the first number you're trying to find.

a₁ is the first number.

d is your scaling value. (n-1) is just there to help you calculate the previous values.

You have the first and last number (the sixth number which is 16). Use those to find your scaling value then it's just plug and chugg after that.

  Kathleen determines the four arithmetic means to be −2.8 , 1.4, 5.6, and 9.8.

Step-by-step explanation:

The overall interval is 14 -(-7) = 21 units in length. The four equally-spaced "means" divide the interval into 5 equal parts, each one of which is ...

  21/5 = 4.2

The four means are ...

  -7 +4.2 = -2.8

  -2.8 +4.2 = 1.4

  1.4 +4.2 = 5.6

  5.6 +4.2 = 9.8

  Kathleen determines the four arithmetic means to be −2.8 , 1.4, 5.6, and 9.8.

Step-by-step explanation:

The overall interval is 14 -(-7) = 21 units in length. The four equally-spaced "means" divide the interval into 5 equal parts, each one of which is ...

  21/5 = 4.2

The four means are ...

  -7 +4.2 = -2.8

  -2.8 +4.2 = 1.4

  1.4 +4.2 = 5.6

  5.6 +4.2 = 9.8

Insert three geometric means between 6 and 96.

Insert three geometric means between 6 and 96.Insert three geometric means between 8 and 5000.

Step-by-step explanation:

this is the correct answer

Insert three geometric means between 6 and 96.

Insert three geometric means between 6 and 96.Insert three geometric means between 8 and 5000.

Step-by-step explanation:

this is the correct answer

Mean.m

fid = fopen('input.txt');

Array = fscanf(fid, '%g');

Amean = AM(Array);

Gmean = GM(Array);

Rmean = RMS(Array);

display('Amlan Das Statistical Package')

display(['arithmetic mean = ', sprintf('%.5f',Amean)]);

disp(['geometric mean = ', sprintf('%.5f',Gmean)]);

disp(['RMS average = ', sprintf('%.5f',Rmean)]);

AM.m

function [ AMean ] = AM( Array )

n = length(Array);

AMean = 0;

for i = 1:n

AMean = AMean + Array(i);

end

AMean = AMean/n;

end

GM.m

function [ GMean ] = GM(Array)

n = length(Array);

GMean = 1;

for i = 1:n

GMean = GMean*Array(i);

end

GMean = GMean^(1/n);

end

RMS.m

function [ RMean ] = RMS( Array)

n = length(Array);

RMean = 0;

for i = 1:n

RMean = RMean + Array(i)^2;

end

RMean = (RMean/n)^(0.5);

end

Mean.m

fid = fopen('input.txt');

Array = fscanf(fid, '%g');

Amean = AM(Array);

Gmean = GM(Array);

Rmean = RMS(Array);

display('Amlan Das Statistical Package')

display(['arithmetic mean = ', sprintf('%.5f',Amean)]);

disp(['geometric mean = ', sprintf('%.5f',Gmean)]);

disp(['RMS average = ', sprintf('%.5f',Rmean)]);

AM.m

function [ AMean ] = AM( Array )

n = length(Array);

AMean = 0;

for i = 1:n

AMean = AMean + Array(i);

end

AMean = AMean/n;

end

GM.m

function [ GMean ] = GM(Array)

n = length(Array);

GMean = 1;

for i = 1:n

GMean = GMean*Array(i);

end

GMean = GMean^(1/n);

end

RMS.m

function [ RMean ] = RMS( Array)

n = length(Array);

RMean = 0;

for i = 1:n

RMean = RMean + Array(i)^2;

end

RMean = (RMean/n)^(0.5);

end

The answer is below

Explanation:

Given that we need to have two or more arithmetic means inserted between two arithmetic extremes, then we have

d= (an - ak) ÷ ( n - k)

For Example: What are the three arithmetic means of the arithmetic extremes 8 and 16, then a1= 12, and a5= 20?

First we use the formula:

d= (an - ak) ÷ (n - k)

= (20 - 12 ) ÷ (5 - 1)

=> (8) ÷ (4) = 2

Our common difference is 2.

Then we can simply:

a5 = 20

a4 = 20 - 2(1) = 18

a3 = 20 - 2(2) = 16

a2 = 20 - 2(3) = 14

a1 = 20 - 2(4) = 12

Now we have our arithmetic sequence 12, 14, 16, 18, 20

Therefore the three arithmetic means of the arithmetic extremes 12 and 20 ==> 14, 16 and 18.