Let the first term a and the common ratio r of a geometric progression be positive integers. If the

Q: Wait, why can't we just use the standard sum of a G.P. formula?

The formula $S_n = \frac{a(r^n - 1)}{r - 1}$ is for the sum of terms, not the sum of squares. Here, we are dealing with $a^2 + (ar)^2 + (ar^2)^2$. Using the standard sum formula would lead you down a completely different path that doesn't account for the squares. Always check what the problem is asking you to sum!

Q: How do I know $a^2$ is the perfect square factor?

Because $a$ is an integer, $a^2$ must be a perfect square. Since $1 + r^2 + r^4$ is also an integer (as $r$ is an integer), $a^2$ must be a divisor of 33033. By finding all perfect square divisors of 33033, we restrict the possible values of $a$ significantly, turning a complex equation into a manageable one.

Q: What if there were multiple perfect square factors?

If there were, you would test each one. For each potential $a$, you would solve $1 + r^2 + r^4 = \frac{33033}{a^2}$. If $r$ turns out to be an integer, you have found a valid solution. In this specific problem, the prime factorization $33033 = 3 \times 7 \times 11^2 \times 13$ makes $11^2$ the most prominent candidate.

Question

Let the first term $a$ and the common ratio $r$ of a geometric progression be positive integers. If the sum of squares of its first three terms is 33033, then the sum of these three terms is equal to

Options:

(A) 241

(B) 231

(C) 210

(D) 220

Sigma Percentile · Accepted Answer

### The Beauty of Integer Constraints

Welcome, fellow explorer of the mathematical universe. Today, we are not just solving a problem; we are embarking on a journey through the elegant world of number theory and geometric progressions.

Imagine standing on the edge of a cliff, looking out at a sequence of numbers that grow with a rhythm—a geometric progression. We have three terms, $a$, $ar$, and $ar^2$, and we are told that the sum of their squares is $33033$.

Our mission is to find the sum of these three terms. The hidden treasure in the problem statement is that $a$ and $r$ are **positive integers**. This constraint is our compass, guiding us toward discrete, countable entities.

### The Algebraic Blueprint

Let us write down what we know. The terms are $a$, $ar$, and $ar^2$. The sum of their squares is given by the equation:

$$a^2 + (ar)^2 + (ar^2)^2 = 33033$$

Expanding this, we get:

$$a^2 + a^2r^2 + a^2r^4 = 33033$$

We can factor out $a^2$ immediately:

$$a^2(1 + r^2 + r^4) = 33033$$

This is the moment where most students panic, seeing one equation and two variables. But remember our compass: $a$ and $r$ are integers. This means $a^2$ must be a perfect square that divides $33033$.

### The Detective Work: Prime Factorization

To find the perfect square factor, we must break $33033$ down to its DNA—its prime factors. Let us perform the factorization:

$$33033 = 3 	imes 11011 = 3 	imes 7 	imes 1573 = 3 	imes 7 	imes 11 	imes 143 = 3 	imes 7 	imes 11^2 	imes 13$$

So, $33033 = 3 	imes 7 	imes 11^2 	imes 13$. Look at that $11^2$! It is a perfect square.

This strongly suggests that $a^2 = 11^2$, which implies $a = 11$. If $a = 11$, then the remaining part of our equation must be:

$$1 + r^2 + r^4 = \frac{33033}{121} = 273$$

### Solving the Quadratic

Now we are left with a beautiful, clean equation in $r$:

$$r^4 + r^2 - 272 = 0$$

This is a quadratic in disguise. Let $x = r^2$. Then $x^2 + x - 272 = 0$.

We need two numbers that multiply to $-272$ and add to $1$. After a moment of reflection, we find $17$ and $-16$. Thus, the equation factors as:

$$(r^2 + 17)(r^2 - 16) = 0$$

Since $r$ is a positive integer, $r^2$ cannot be $-17$. Therefore, $r^2 = 16$, which means $r = 4$.

### The Final Victory

With $a = 11$ and $r = 4$, our terms are $11$, $11 	imes 4 = 44$, and $11 	imes 16 = 176$. The question asks for the sum of these three terms:

$$Sum = 11 + 44 + 176 = 231$$

We have arrived at the destination. The beauty of this problem lies not in the calculation, but in the realization that constraints are not limitations—they are the keys that unlock the solution. The final answer is **231**.

Animated Solution for Mathematics - Sequence and Series: Let the first term $a$ and the common ratio $r$ of a geometric progression be positive integers. If the sum of squares of its first three terms is 33033, then the sum of these three terms is equal to

Select Answer:

Visualized Solution

Problem Setup and Variables

Formulating the Equation

Factoring the Expression

Prime Factorization of $33033$

Comparing Both Sides

Solving for $r$

Finding the Common Ratio

Calculating the Terms

Final Sum Calculation

The Sigma Insight: Geometric Progression (G.P.)

The Beauty of Integer Constraints

The Algebraic Blueprint

The Detective Work

Solving the Quadratic

The Final Victory

Similar Questions

The sum of first four terms of a geometric progression (G.P.) is $\frac{65}{12}$ and the sum of their respective reciprocals is $\frac{65}{18}$ . If the product of first three terms of the G.P. is 1, and the third term is $α$ , then $2 α$ is

In a geometric progression consisting of positive terms, each term equals the sum of the next two terms. Then the common ratio of its progression is equals

Three numbers are in an increasing geometric progression with common ratio $r$ . If the middle number is doubled, then the new numbers are in an arithmetic progression with common difference $d$ . If the fourth term of $GP$ is $3 r^{2}$ , then $r^{2} - d$ is equal to :

If the sum of the second, third and fourth terms of a positive term G.P. is 3 and the sum of its sixth, seventh and.eighth terms is 243, then the sum of the first 50 terms of this G.P. is :

Let $A_{1}, A_{2}, A_{3}, \dots$ be an increasing geometric progression of positive real numbers. If $A_{1} A_{3} A_{5} A_{7} = \frac{1}{1296}$ and $A_{2} + A_{4} = \frac{7}{36}$ , then, the value of $A_{6} + A_{8} + A_{10}$ is equal to

If the sum of the second, fourth and sixth terms of a G.P. of positive terms is 21 and the sum of its eighth, tenth and twelfth terms is 15309, then the sum of its first nine terms is :

In an increasing, geometric series, the sum of the second and the sixth term is $\frac{25}{2}$ and the product of the third and fifth term is 25. Then, the sum of $4^{t h}, 6^{t h}$ and $8^{t h}$ terms is equal to :

The third term of a geometric progression is $4$ . The product of the first five terms is

If the sum and product of four positive consecutive terms of a G.P., are 126 and 1296, respectively, then the sum of common ratios of all such GPs is