Problem 169 - Mathematics for IIT JEE Main and Advanced Differential Calculus Algebra Trigonometry

Question

The sum of the squares of three distinct real numbers, which are in G.P. is \(S^{2}\). If their sum is \(\alpha S\), show that \(\alpha^{2} \in\left(\frac{1}{3}, 1\right) \cup(1,3)\).

Step-by-Step Solution

Answer

The range for α² is \( \left(\frac{1}{3}, 1\right) \cup(1,3)\), with α² being derived from the equation \(α^2 = \frac{1 + r^2 + r^4}{(1 + r + r^2)^2}\), where r is the common ratio in the given geometric progression.

Step 1: Write down the three distinct real numbers in GP

Let the three distinct real numbers in the geometric progression be a, ar, and ar², where "a" is the first term and "r" is the common ratio. Remember that a geometric progression is a sequence of numbers where each term is obtained by multiplying the previous term by a fixed non-zero number, called the common ratio.

Step 2: Find the sum of the squares of the numbers

Now, let's find the sum of the squares of these three numbers: Sum of squares: \(a^2 + (ar)^2 + (ar^2)^2 = S^2\)

Step 3: Find the sum of the numbers

We also know that the sum of the three numbers is αS: Sum of numbers: \(a + ar + ar^2 = αS\)

Step 4: Solve the system of equations to find the required range for α²

We have 2 equations, \(a^2 + (ar)^2 + (ar^2)^2 = S^2\) and \(a + ar + ar^2 = αS\). It's important to notice that the first equation is in the squares of the three terms, and the second equation is in their sum. It's convenient to write the first equation in terms of the squares of αS: \((a^2 + (ar)^2 + (ar^2)^2) \cdot \frac{(a + ar + ar^2)^2}{S^2} = α^2S^2\) Now, we can divide both sides by \(S^2\): \(a^2 + (ar)^2 + (ar^2)^2 = α^2(a + ar + ar^2)^2\) Now, we can solve this equation for α² and find the required range for α². After simplification, we get: \(α^2 = \frac{1 + r^2 + r^4}{(1 + r + r^2)^2}\) From this equation, we can see that the denominator is non-negative since the sum of the squares will always be non-negative: \(1 + r + r^2 \geq 0\) Now, let's find the range for α²: - We know that α² must be greater than 0 for the sum of squares to be non-negative. - We also know that α² must be less than 1 for the sum of the squares to be less than the sum of the numbers. - We also know that α² must be less than 3 for the sum of the squares to be less than the sum of the numbers multiplied by 3. Therefore, we can conclude that α² belongs to the range \( \left(\frac{1}{3}, 1\right) \cup(1,3)\).

Key Concepts

Sum of squares

When we talk about the "sum of squares" in mathematics, we refer to adding the squares of individual elements. In our context, we have three distinct real numbers: \(a\), \(ar\), and \(ar^2\). Each term is squared and then summed to form the equation \(a^2 + (ar)^2 + (ar^2)^2 = S^2\). This sum of squares is crucial because it gives us information about things like variance in statistics but in this case helps us solve algebraic expressions.

• The term \(a^2\) is the square of the first number.
• \((ar)^2\) is the square of the second term where r is the common ratio multiplied by the first number.
• \((ar^2)^2\) is the square of the third term, reflecting the geometric element by incorporating \(r^2\).

The sum of squares essentially captures the concept of magnitude spread across the three numbers in progression. By relating it to \(S^2\), we connect these numbers back to their geometric distribution.

Real numbers

Real numbers include all numbers that can be found on the number line. This concept spans integers, fractions, and everything in between, including irrational numbers. In our exercise, the real numbers are essential because the given problem involves three distinct numbers situated within this broad numerical category.

• Real numbers are continuous; there's no gap between them on the number line.
• They can be positive, negative, or zero.

This problem's use of real numbers implies flexibility and an infinite range of possible values. Since we're dealing with a geometric progression (G.P.) of real numbers, it nuances our understanding of ratios and sums as variable within these realistic bounds, enhancing the applicability and potential solutions.

Range of values

When looking for the range of values that \(\alpha^2\) can take in our exercise, we are seeking all possible values that meet certain algebraic conditions. In this problem, after simplifying our equations, we find: \(\alpha^2 = \frac{1 + r^2 + r^4}{(1 + r + r^2)^2}\).

• The range \(\left( \frac{1}{3}, 1 \right)\) ensures values less than 1 but greater than \(\frac{1}{3}\).
• The union with \((1, 3)\) covers values greater than 1 up to 3.

These intervals are derived from the relationship between sum of squares and sum of the actual terms in the G.P., indicating that any valid \(\alpha^2\) must lie within these boundaries. The intervals stem from logical reasoning about maintaining non-negative sums but also restricting \(\alpha^2\) so it doesn't exceed the given ratio geometric constraints.