Problem 81 - A Complete Resource Book in Mathematics for JEE Main 2017

Question

The three successive terms of a G.P. will form the sides of a triangle if the common ratio \(r\) satisfies the inequality(A) \(\frac{\sqrt{3}-1}{2}

Step-by-Step Solution

Answer

Option (B) is correct: \( \frac{\sqrt{5}-1}{2}

Step 1: Define the Terms and Condition

Let the three terms of the G.P. be \( a \), \( ar \), \( ar^2 \). For these three terms to form a triangle, the sum of any two sides must be greater than the third side. This gives the triangle inequality properties: \( a + ar > ar^2 \), \( a + ar^2 > ar \), and \( ar + ar^2 > a \).

Step 2: Simplify the First Inequality

The first inequality is \( a + ar > ar^2 \). Factoring out \( a \), we have \( a(1 + r) > ar^2 \). Dividing both sides by \( a \), we get \( 1 + r > r^2 \).

Step 3: Rearrange the First Inequality

Rearrange the inequality \( 1 + r > r^2 \) to \( r^2 - r - 1 < 0 \). This is a quadratic inequality that needs to be solved to find the values of \( r \).

Step 4: Solve the Quadratic Equation

Solve the equation \( r^2 - r - 1 = 0 \) to find the roots. Using the quadratic formula, \( r = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), where \( a = 1 \), \( b = -1 \), \( c = -1 \). This gives the roots \( r = \frac{1 \pm \sqrt{5}}{2} \).

Step 5: Determine the Range of r

The solutions \( \frac{1 - \sqrt{5}}{2} \) and \( \frac{1 + \sqrt{5}}{2} \) determine when the inequality \( r^2 - r - 1 < 0 \) holds. The inequality is satisfied for \( \frac{1 - \sqrt{5}}{2} < r < \frac{1 + \sqrt{5}}{2} \). Since \( \frac{1 - \sqrt{5}}{2} \) is negative, we only consider the positive range: \( 0 < r < \frac{1 + \sqrt{5}}{2} \).

Step 6: Check which Answer Option Fits

Compare the derived range for the inequality \( \frac{1 - \sqrt{5}}{2} < r < \frac{1 + \sqrt{5}}{2} \) with the provided options. The correct range matches option (B), i.e., \( \frac{\sqrt{5}-1}{2} < r < \frac{\sqrt{5}+1}{2} \).

Key Concepts

Triangle Inequality

The Triangle Inequality is a fundamental principle in geometry. It states that for any three side lengths to form a triangle, the sum of the lengths of any two sides must be greater than the length of the third side. This ensures that the sides can "close" to form a closed shape, which is a triangle.
To illustrate, consider sides of lengths: \( a \), \( b \), and \( c \). The inequalities can be expressed as:

• \( a + b > c \)
• \( a + c > b \)
• \( b + c > a \)

These conditions ensure the feasibility of forming a triangle with the given sides. If any of these inequalities are not met, the sides would not "meet" to form a triangle, indicating a geometrically impossible situation. In the context of a Geometric Progression (G.P.), the sides coalescing from successive terms must also satisfy these inequalities to be considered viable sides of a triangle.

Quadratic Inequality

A quadratic inequality involves a quadratic expression which needs to be solved to find ranges or intervals. Solving these inequalities involves understanding how the quadratic equation behaves with different values:
Let's consider a quadratic inequality in the form of \( ax^2 + bx + c < 0 \). Here's how to solve it:

• First, find the roots of the corresponding quadratic equation \( ax^2 + bx + c = 0 \). This can be done using the quadratic formula, \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \).
• Utilize these roots to divide the number line into intervals. You'll check the sign of the quadratic in each interval to see where the inequality is satisfied.
• The inequality's solution will usually lie between the roots where the parabola opens upwards and below the x-axis. Conversely, for \( ax^2 + bx + c > 0 \), the solution will be outside these roots.

Solving quadratic inequalities often involves testing solutions within intervals and determining the range of values that satisfy the inequality. For inequalities like \( r^2 - r - 1 < 0 \), the solution region is where the parabola dips below the x-axis, providing potential side lengths for our triangle problem.

Roots of a Quadratic Equation

Finding the roots of a quadratic equation is essential in understanding and solving problems related to quadratic inequalities. The roots determine the points where the quadratic function crosses the x-axis.
A quadratic equation takes the form \( ax^2 + bx + c = 0 \). The roots can be found using the quadratic formula:
\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
Here, \( b^2 - 4ac \) is called the discriminant, which determines the nature of the roots. If:

• The discriminant is positive, there are two distinct real roots.
• It is zero, indicating one real root (a repeated root).
• It is negative, suggesting two complex roots.

The roots play a crucial role in defining the intervals of inequality solutions. For the equation \( r^2 - r - 1 = 0 \), the roots were determined to be \( \frac{1 \pm \sqrt{5}}{2} \). These are used to understand the behavior of the inequality \( r^2 - r - 1 < 0 \) and are key to establishing the valid range for the common ratio \( r \) of our geometric progression when they serve as triangle sides.

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Problem 80

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Problem 82

Other exercises in this chapter

Problem 79

The largest term of the sequence \(\frac{1}{503}, \frac{4}{524}, \frac{9}{581}, \frac{16}{692}, \ldots\) is (A) \(\frac{16}{692}\) (B) \(\frac{4}{524}\) (C) \(\

View solution

Problem 80

The coefficient of \(x^{99}\) and \(x^{98}\) in the polynomial \((x-1)(x-2)(x-3) \ldots(x-100)\) are (A) \(-5050\) and 12482075 (B) \(-4050\) and 12582075 (C) \

View solution

Problem 82

If the sides of a right angled triangle are in G.P., then the cosine of the greater acute angle is (A) \(\frac{1}{1+\sqrt{5}}\) (B) \(\frac{1}{1-\sqrt{5}}\) (C)

View solution

Problem 84

If the \(p\) th, \(q\) th and \(r\) th terms of both an A.P. and a G.P. be respectively \(a, b\) and \(c\), then (A) \(a^{c} \cdot c^{b} \cdot b^{a}=a^{c} \cdot

View solution

Key Concepts

Triangle Inequality

The Triangle Inequality is a fundamental principle in geometry. It states that for any three side lengths to form a triangle, the sum of the lengths of any two sides must be greater than the length of the third side. This ensures that the sides can "close" to form a closed shape, which is a triangle.
To illustrate, consider sides of lengths: \( a \), \( b \), and \( c \). The inequalities can be expressed as:

• \( a + b > c \)
• \( a + c > b \)
• \( b + c > a \)

These conditions ensure the feasibility of forming a triangle with the given sides. If any of these inequalities are not met, the sides would not "meet" to form a triangle, indicating a geometrically impossible situation. In the context of a Geometric Progression (G.P.), the sides coalescing from successive terms must also satisfy these inequalities to be considered viable sides of a triangle.

Quadratic Inequality

A quadratic inequality involves a quadratic expression which needs to be solved to find ranges or intervals. Solving these inequalities involves understanding how the quadratic equation behaves with different values:
Let's consider a quadratic inequality in the form of \( ax^2 + bx + c < 0 \). Here's how to solve it:

• First, find the roots of the corresponding quadratic equation \( ax^2 + bx + c = 0 \). This can be done using the quadratic formula, \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \).
• Utilize these roots to divide the number line into intervals. You'll check the sign of the quadratic in each interval to see where the inequality is satisfied.
• The inequality's solution will usually lie between the roots where the parabola opens upwards and below the x-axis. Conversely, for \( ax^2 + bx + c > 0 \), the solution will be outside these roots.

Solving quadratic inequalities often involves testing solutions within intervals and determining the range of values that satisfy the inequality. For inequalities like \( r^2 - r - 1 < 0 \), the solution region is where the parabola dips below the x-axis, providing potential side lengths for our triangle problem.

Roots of a Quadratic Equation

Finding the roots of a quadratic equation is essential in understanding and solving problems related to quadratic inequalities. The roots determine the points where the quadratic function crosses the x-axis.
A quadratic equation takes the form \( ax^2 + bx + c = 0 \). The roots can be found using the quadratic formula:
\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
Here, \( b^2 - 4ac \) is called the discriminant, which determines the nature of the roots. If:

• The discriminant is positive, there are two distinct real roots.
• It is zero, indicating one real root (a repeated root).
• It is negative, suggesting two complex roots.

The roots play a crucial role in defining the intervals of inequality solutions. For the equation \( r^2 - r - 1 = 0 \), the roots were determined to be \( \frac{1 \pm \sqrt{5}}{2} \). These are used to understand the behavior of the inequality \( r^2 - r - 1 < 0 \) and are key to establishing the valid range for the common ratio \( r \) of our geometric progression when they serve as triangle sides.