Key Concepts and Formulas

1. Arithmetic Progression (AP): For three numbers $x, y, z$ in AP, the middle term is the average of the other two: $2y = x+z$.
2. Geometric Progression (GP): For three numbers $x, y, z$ in GP, the square of the middle term is the product of the other two: $y^2 = xz$.
3. Arithmetic Mean (AM): The sum of $n$ numbers divided by $n$. For $a, b, c$, AM = $\frac{a+b+c}{3}$.
4. Geometric Mean (GM): The $n$-th root of the product of $n$ numbers. For $a, b, c$, GM = $\sqrt[3]{abc}$. The cube of the GM is $(\sqrt[3]{abc})^3 = abc$.

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Step-by-Step Solution

Step 1: Formulate Equations from the Given Information

We are given three real numbers $a, b, c$.

• $a, b, c$ are in arithmetic progression: $2b = a+c \quad \text{....(1)}$
• $a+1, b, c+3$ are in geometric progression: $b^2 = (a+1)(c+3) \quad \text{....(2)}$
• The arithmetic mean of $a, b, c$ is 8: $\frac{a+b+c}{3} = 8 \quad \text{....(3)}$
• We are also given that $a > 10$.

Step 2: Determine the Value of 'b'

From Equation (3), we can find the sum of $a, b, c$: $a+b+c = 3 \times 8$ $a+b+c = 24 \quad \text{....(4)}$ Now, substitute the expression for $a+c$ from Equation (1) into Equation (4): $(2b) + b = 24$ $3b = 24$ $b = 8$

Step 3: Simplify Equations (1) and (2) using the Value of 'b'

Substitute $b=8$ into Equation (1): $2(8) = a+c$ $16 = a+c \quad \text{....(5)}$ Substitute $b=8$ into Equation (2): $8^2 = (a+1)(c+3)$ $64 = (a+1)(c+3)$ Expand the right side: $64 = ac + 3a + c + 3$ Rearrange the terms: $ac + 3a + c = 61 \quad \text{....(6)}$

Step 4: Solve the System of Equations for 'a' and 'c'

We have a system of two equations:

1. $a+c = 16$
2. $ac + 3a + c = 61$

From Equation (5), express $c$ in terms of $a$: $c = 16-a \quad \text{....(7)}$ Substitute this expression for $c$ into Equation (6): $a(16-a) + 3a + (16-a) = 61$ $16a - a^2 + 3a + 16 - a = 61$ Combine like terms: $-a^2 + (16a+3a-a) + 16 = 61$ $-a^2 + 18a + 16 = 61$ Move all terms to one side to form a quadratic equation: $-a^2 + 18a + 16 - 61 = 0$ $-a^2 + 18a - 45 = 0$ Multiply by $-1$: $a^2 - 18a + 45 = 0$ Factor the quadratic equation: $(a-15)(a-3) = 0$ This yields two possible values for $a$: $a=15$ or $a=3$.

Step 5: Apply the Condition $a > 10$ and Find 'c'

The problem states that $a > 10$. Therefore, we choose $a=15$ and reject $a=3$. Now, use Equation (7) to find $c$: $c = 16 - a$ $c = 16 - 15$ $c = 1$ The three numbers are $a=15, b=8, c=1$.

Step 6: Calculate the Cube of the Geometric Mean

The geometric mean of $a, b, c$ is $\sqrt[3]{abc}$. The cube of the geometric mean is $(\sqrt[3]{abc})^3 = abc$. Substitute the values of $a, b,$ and $c$: $abc = 15 \times 8 \times 1$ $abc = 120$

Common Mistakes & Tips

• Algebraic Errors: Be careful with expanding expressions like $(a+1)(c+3)$ and when rearranging terms in quadratic equations.
• Ignoring Conditions: Always check if the derived values satisfy all given conditions, especially inequalities like $a>10$.
• Direct Substitution: Using the AM of $a, b, c$ and the AP condition ($2b=a+c$) allows for a direct calculation of $b$, which simplifies the problem significantly.

Summary

The problem involved translating conditions on arithmetic and geometric progressions into algebraic equations. By utilizing the given arithmetic mean, we efficiently found the value of $b$. Subsequently, we solved a system of equations for $a$ and $c$, applying the constraint $a>10$ to select the correct value for $a$. Finally, we calculated the product $abc$, which represents the cube of the geometric mean.

The final answer is \boxed{120}.