Understanding the Problem

The question provides information about the arithmetic mean (AM), geometric mean (GM), and median of six positive numbers: a, a, b, b, c, c, with the condition that a < b < c. We are given the values of the AM, GM, and median and asked to find the sum of the squares of these six numbers, which is \(2a^2 + 2b^2 + 2c^2\).

Let's use the given information to find the values of a, b, and c.

Using the Median Information

The six numbers in non-decreasing order are a, a, b, b, c, c because a < b < c. For a set of 6 numbers, the median is the average of the 3rd and 4th numbers in the ordered list.

The 3rd number is b.

The 4th number is b.

The median is \(\frac{b + b}{2} = \frac{2b}{2} = b\).

We are given that the median is 2. Therefore, we have our first value:

\(b = 2\)

Using the Arithmetic Mean (AM) Information

The AM of six numbers is the sum of the numbers divided by 6.

AM = \(\frac{a + a + b + b + c + c}{6} = \frac{2a + 2b + 2c}{6} = \frac{2(a + b + c)}{6} = \frac{a + b + c}{3}\)

We are given that the AM is \(\frac{7}{3}\).

So, \(\frac{a + b + c}{3} = \frac{7}{3}\)

Multiplying both sides by 3 gives:

\(a + b + c = 7\)

Substitute the value of b = 2 into this equation:

\(a + 2 + c = 7\)

Subtracting 2 from both sides gives a relationship between a and c:

\(a + c = 5\)

Using the Geometric Mean (GM) Information

The GM of six positive numbers is the 6th root of their product.

GM = \(\sqrt[6]{a \cdot a \cdot b \cdot b \cdot c \cdot c} = \sqrt[6]{a^2 b^2 c^2} = \sqrt[6]{(abc)^2}\)

Using the property \(\sqrt[n]{x^m} = x^{m/n}\), we get:

GM = \((abc)^{2/6} = (abc)^{1/3} = \sqrt[3]{abc}\)

We are given that the GM is 2.

So, \(\sqrt[3]{abc} = 2\)

Cube both sides of the equation:

\((\sqrt[3]{abc})^3 = 2^3\)

\(abc = 8\)

Substitute the value of b = 2 into this equation:

\(a \cdot 2 \cdot c = 8\)

Dividing both sides by 2 gives a relationship between a and c:

\(ac = 4\)

Solving for a and c

We have a system of two equations with two variables, a and c:

1. \(a + c = 5\)
2. \(ac = 4\)

From equation (1), we can express c as \(c = 5 - a\). Substitute this into equation (2):

\(a(5 - a) = 4\)

\(5a - a^2 = 4\)

Rearrange the terms to form a quadratic equation:

\(a^2 - 5a + 4 = 0\)

Factor the quadratic equation:

\((a - 1)(a - 4) = 0\)

This gives two possible values for a: \(a = 1\) or \(a = 4\).

Checking the Condition a < b < c

We know b = 2. We need to check which pair of (a, c) satisfies the condition a < b < c.

• If \(a = 1\): Using \(a + c = 5\), we get \(1 + c = 5\), so \(c = 4\). The values are a = 1, b = 2, c = 4. Check the condition: 1 < 2 < 4. This is true.
• If \(a = 4\): Using \(a + c = 5\), we get \(4 + c = 5\), so \(c = 1\). The values are a = 4, b = 2, c = 1. Check the condition: 4 < 2 < 1. This is false (4 is not less than 2, and 2 is not less than 1).

Therefore, the only valid solution is a = 1, b = 2, and c = 4.

Calculating the Sum of Squares

The six numbers are a, a, b, b, c, c. With a = 1, b = 2, and c = 4, the numbers are 1, 1, 2, 2, 4, 4.

The sum of the squares of these six numbers is \(a^2 + a^2 + b^2 + b^2 + c^2 + c^2 = 2a^2 + 2b^2 + 2c^2\).

Substitute the values of a, b, and c:

Sum of squares = \(2(1)^2 + 2(2)^2 + 2(4)^2\)

Sum of squares = \(2(1) + 2(4) + 2(16)\)

Sum of squares = \(2 + 8 + 32\)

Sum of squares = \(42\)

Summary of Results

Using \(b=2\) in the AM and GM equations gave us \(a+c=5\) and \(ac=4\). Solving these simultaneously with the condition \(a < b < c\) (i.e., \(a < 2 < c\)) led to \(a=1\) and \(c=4\).

The numbers are 1, 1, 2, 2, 4, 4.

The sum of squares is \(1^2 + 1^2 + 2^2 + 2^2 + 4^2 + 4^2 = 1 + 1 + 4 + 4 + 16 + 16 = 42\).

The sum of the squares of all six numbers is 42.