Key Concepts and Formulas

• Geometric Progression (GP): For three consecutive terms $x, y, z$ in GP, the middle term is the geometric mean of the other two, satisfying $y^2 = xz$.
• Arithmetic Progression (AP): For three consecutive terms $x, y, z$ in AP, the middle term is the arithmetic mean of the other two, satisfying $2y = x + z$.
• Solving Quadratic Equations: The quadratic formula $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ is used to find the roots of an equation of the form $ax^2 + bx + c = 0$.

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Step-by-Step Solution

Step 1: Formulate equations from the given GP condition. We are given that $a, b, \frac{1}{18}$ are in a geometric progression. Using the property of GP, the square of the middle term is equal to the product of the other two terms: $b^2 = a \times \frac{1}{18}$ This gives us the first equation relating $a$ and $b$: $a = 18b^2 \quad \text{(Equation 1)}$

• Reasoning: This step translates the information about the geometric progression into an algebraic relationship between $a$ and $b$.

Step 2: Formulate equations from the given AP condition. We are given that $\frac{1}{a}, 10, \frac{1}{b}$ are in an arithmetic progression. Using the property of AP, twice the middle term is equal to the sum of the other two terms: $2 \times 10 = \frac{1}{a} + \frac{1}{b}$ This simplifies to: $20 = \frac{1}{a} + \frac{1}{b} \quad \text{(Equation 2)}$

• Reasoning: This step translates the information about the arithmetic progression into another algebraic relationship between $a$ and $b$.

Step 3: Substitute Equation 1 into Equation 2 to solve for $b$. Substitute $a = 18b^2$ from Equation 1 into Equation 2: $20 = \frac{1}{18b^2} + \frac{1}{b}$ To eliminate the denominators, multiply the entire equation by $18b^2$: $20 \times (18b^2) = \left(\frac{1}{18b^2}\right) \times (18b^2) + \left(\frac{1}{b}\right) \times (18b^2)$ $360b^2 = 1 + 18b$ Rearrange this into a standard quadratic equation: $360b^2 - 18b - 1 = 0$

• Reasoning: By substituting the expression for $a$ from the GP condition into the AP condition, we obtain a single equation with only one variable, $b$, which can then be solved.

Step 4: Solve the quadratic equation for $b$. Using the quadratic formula $b = \frac{-B \pm \sqrt{B^2 - 4AC}}{2A}$ with $A=360$, $B=-18$, and $C=-1$: $b = \frac{-(-18) \pm \sqrt{(-18)^2 - 4(360)(-1)}}{2(360)}$ $b = \frac{18 \pm \sqrt{324 + 1440}}{720}$ $b = \frac{18 \pm \sqrt{1764}}{720}$ The square root of $1764$ is $42$. $b = \frac{18 \pm 42}{720}$ This yields two possible values for $b$: $b_1 = \frac{18 + 42}{720} = \frac{60}{720} = \frac{1}{12}$ $b_2 = \frac{18 - 42}{720} = \frac{-24}{720} = -\frac{1}{30}$ Since the problem states that $a$ and $b$ are positive numbers, we choose the positive value for $b$: $b = \frac{1}{12}$

• Reasoning: The quadratic formula provides the solutions to the quadratic equation. The condition that $a$ and $b$ are positive allows us to select the valid solution for $b$.

Step 5: Calculate the value of $a$ using Equation 1. Substitute the value of $b = \frac{1}{12}$ into Equation 1 ($a = 18b^2$): $a = 18 \times \left(\frac{1}{12}\right)^2$ $a = 18 \times \frac{1}{144}$ $a = \frac{18}{144}$ Simplifying the fraction: $a = \frac{1}{8}$

• Reasoning: With the value of $b$ determined, we can now find the value of $a$ using the relationship established in Step 1.

Step 6: Calculate the required expression $16a + 12b$. Substitute the values $a = \frac{1}{8}$ and $b = \frac{1}{12}$ into the expression $16a + 12b$: $16a + 12b = 16 \left(\frac{1}{8}\right) + 12 \left(\frac{1}{12}\right)$ $16a + 12b = 2 + 1$ $16a + 12b = 3$

• Reasoning: This is the final step where we substitute the found values of $a$ and $b$ into the expression requested by the problem to obtain the final answer.

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Common Mistakes & Tips

• Ensure you correctly apply the properties of GP ($y^2 = xz$) and AP ($2y = x+z$) to the given terms.
• Pay close attention to algebraic manipulations, especially when dealing with fractions and solving the quadratic equation to avoid arithmetic errors.
• Always use the given conditions (e.g., $a, b$ are positive) to discard extraneous solutions.
• Before concluding, it's a good practice to verify if the calculated $a$ and $b$ satisfy the original conditions of the problem.

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Summary

The problem required us to use the properties of geometric and arithmetic progressions to set up a system of two equations with two variables, $a$ and $b$. By substituting one equation into the other, we obtained a quadratic equation in $b$, which we solved using the quadratic formula. The condition that $a$ and $b$ are positive helped us select the correct value for $b$. Subsequently, we found the value of $a$ and then computed the required expression $16a + 12b$.

The final answer is $\boxed{3}$.