Key Concepts and Formulas

• Geometric Progression (GP): A sequence where each term after the first is found by multiplying the previous one by a fixed, non-zero number called the common ratio ($r$). The terms are $\alpha, \alpha r, \alpha r^2, \dots$.
• Sum of Squares: The problem involves the sum of the squares of the first three terms: $\alpha^2 + (\alpha r)^2 + (\alpha r^2)^2$.
• Prime Factorization: Essential for breaking down large numbers to identify perfect square factors.

Step-by-Step Solution

Step 1: Set up the equation from the problem statement. The problem states that the first term ($\alpha$) and the common ratio ($r$) of a geometric progression are positive integers. The sum of the squares of its first three terms is 33033. The first three terms are $\alpha$, $\alpha r$, and $\alpha r^2$. Their squares are $\alpha^2$, $(\alpha r)^2 = \alpha^2 r^2$, and $(\alpha r^2)^2 = \alpha^2 r^4$. The sum of these squares is given as 33033: $\alpha^2 + \alpha^2 r^2 + \alpha^2 r^4 = 33033$

Step 2: Factor the equation and prime factorize the constant. Factor out $\alpha^2$ from the left side of the equation: $\alpha^2 (1 + r^2 + r^4) = 33033$ Since $\alpha$ is a positive integer, $\alpha^2$ must be a perfect square. We need to find the prime factorization of 33033 to identify any perfect square factors. $33033 = 3 \times 11011$ $11011 = 7 \times 1573$ $1573 = 11 \times 143$ $143 = 11 \times 13$ So, the prime factorization of 33033 is $3 \times 7 \times 11^2 \times 13$. The equation becomes: $\alpha^2 (1 + r^2 + r^4) = 3 \times 7 \times 11^2 \times 13$

Step 3: Determine the value of $\alpha$. From the prime factorization $3 \times 7 \times 11^2 \times 13$, the only perfect square factor is $11^2$. Since $\alpha^2$ must be a perfect square, we can equate $\alpha^2$ to $11^2$. $\alpha^2 = 11^2$ As $\alpha$ is a positive integer, we take the positive square root: $\alpha = 11$

Step 4: Determine the value of $r$. Substitute $\alpha^2 = 11^2 = 121$ back into the factored equation: $121 (1 + r^2 + r^4) = 33033$ Divide both sides by 121: $1 + r^2 + r^4 = \frac{33033}{121}$ $1 + r^2 + r^4 = 273$ Rearrange the equation to solve for $r$: $r^4 + r^2 - 272 = 0$ This is a quadratic equation in terms of $r^2$. Let $x = r^2$. The equation becomes: $x^2 + x - 272 = 0$ We can factor this quadratic equation. We need two numbers that multiply to -272 and add to 1. These numbers are 17 and -16. $(x + 17)(x - 16) = 0$ Substitute back $x = r^2$: $(r^2 + 17)(r^2 - 16) = 0$ This gives two possibilities for $r^2$:

1. $r^2 + 17 = 0 \implies r^2 = -17$. This is not possible for a real number $r$, and hence not for a positive integer $r$.
2. $r^2 - 16 = 0 \implies r^2 = 16$.

Taking the square root of $r^2 = 16$, we get $r = \pm 4$. Since the problem states that $r$ is a positive integer, we choose $r = 4$.

Step 5: Calculate the sum of the first three terms. We have found $\alpha = 11$ and $r = 4$. The first three terms of the GP are: First term: $\alpha = 11$ Second term: $\alpha r = 11 \times 4 = 44$ Third term: $\alpha r^2 = 11 \times 4^2 = 11 \times 16 = 176$ The sum of these three terms is: $\text{Sum} = 11 + 44 + 176$ $\text{Sum} = 55 + 176$ $\text{Sum} = 231$

Common Mistakes & Tips

• Integer Constraints: Always pay close attention to the conditions that $\alpha$ and $r$ are positive integers. This helps in discarding invalid solutions (e.g., negative square roots, non-integer values).
• Prime Factorization Accuracy: Ensure your prime factorization is correct, as it's the foundation for identifying $\alpha^2$.
• Solving Quadratic Equations: Recognize equations that can be treated as quadratic in a substituted variable (like $r^2$) for efficient solving.

Summary

The problem required us to set up an equation for the sum of squares of the first three terms of a GP, using the given information that the first term and common ratio are positive integers. By prime factorizing the given sum, we identified the value of $\alpha^2$ and thus $\alpha$. Substituting this back allowed us to form a quadratic equation in $r^2$, which we solved to find the value of $r$. Finally, we calculated the sum of the first three terms using the determined values of $\alpha$ and $r$.

The sum of these three terms is 231.

The final answer is $\boxed{\text{231}}$.