If three successive terms of a G.P. with common ratio r(r 1) are the lengths of the sides of a tri

Q: Wait, why do we only check $a + ar > ar^2$? Shouldn't we check all three triangle inequalities?

That is a fantastic question! In a general triangle with sides $x, y, z$, you would indeed need to check $x+y > z$, $x+z > y$, and $y+z > x$. However, because we know our sides are $a, ar, ar^2$ and $r > 1$, we already know $ar^2$ is the largest side. If the sum of the two smaller sides is greater than the largest side, the other two inequalities are automatically satisfied. It is a beautiful shortcut that saves us time!

Q: Why is the greatest integer of $-r$ equal to $-2$?

This is a classic trap! The greatest integer function $[x]$ returns the largest integer less than or equal to $x$. If $1 < r < 1.618$, then multiplying by $-1$ flips the inequality to $-1.618 < -r < -1$. On the number line, the integers to the left of $-1.618$ are $-2, -3, \dots$. The largest of these is $-2$. So, $[-r] = -2$.

Q: How do we know $a$ is positive? Could it be negative?

In the context of geometry, $a$ represents the length of a side of a triangle. By definition, a length must be a positive real number ($a > 0$). If $a$ were zero or negative, it wouldn't represent a physical side length, and the triangle wouldn't exist. So, we are perfectly safe to divide by $a$!

Question

If three successive terms of a G.P. with common ratio $r(r > 1)$ are the lengths of the sides of a triangle and $[r]$ denotes the greatest integer less than or equal to $r$, then $3[r] + [-r]$ is equal to :

Sigma Percentile · Accepted Answer

### Analyzing the Setup Imagine you are standing on the edge of a mathematical landscape where three lengths, $a$, $ar$, and $ar^2$, are the physical sides of a triangle. This is the world of Geometric Progressions, where growth is constant and predictable. We are given that the common ratio $r$ is strictly greater than one ($r > 1$). This condition implies that the sides are ordered as $a < ar < ar^2$, making $ar^2$ the longest side. ### The Gatekeeper: The Triangle Inequality For these three segments to form a closed triangle, they must satisfy the **Triangle Inequality**. This theorem dictates that the sum of any two sides must be strictly greater than the third. Because $ar^2$ is the longest side, we only need to ensure that the sum of the two smaller sides, $a$ and $ar$, is greater than the longest side: $$a + ar > ar^2$$ Since $a$ is a physical length, $a > 0$. Dividing the inequality by $a$ yields the elegant expression: $$1 + r > r^2$$ ### The Quadratic Challenge To solve $1 + r > r^2$, we rearrange the terms into a standard quadratic form: $$r^2 - r - 1 < 0$$ To find the region where the parabola dips below the $x$-axis, we first identify the roots of the equation $r^2 - r - 1 = 0$ using the quadratic formula: $$r = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$ Substituting the coefficients, we find the roots to be: $$r = \frac{1 \pm \sqrt{5}}{2}$$ The root $\phi = \frac{1 + \sqrt{5}}{2}$ is the famous **Golden Ratio**, approximately $1.618$. Since the quadratic must be less than zero, $r$ must lie between the two roots. ### The Final Intersection We must now intersect our calculated range with the initial constraint $r > 1$. The valid range for $r$ is: $$1 < r < \frac{1 + \sqrt{5}}{2}$$ This implies $1 < r < 1.618$. We now evaluate the greatest integer functions based on this interval. For $[r]$, since $r$ is between $1$ and $1.618$, the greatest integer less than or equal to $r$ is: $$[r] = 1$$ For $[-r]$, we multiply the range by $-1$ to get $-1.618 < -r < -1$. The greatest integer less than or equal to $-1.618$ is: $$[-r] = -2$$ ### Final Calculation Substituting these values into the expression $3[r] + [-r]$, we obtain: $$3(1) + (-2) = 3 - 2 = 1$$ The final result is **1**.

Animated Solution for Mathematics - Sequence and Series: If three successive terms of a G.P. with common ratio $r (r > 1)$ are the lengths of the sides of a triangle and $[r]$ denotes the greatest integer less than or equal to $r$ , then $3 [r] + [- r]$ is equal to :

Enter Numerical Value:

Visualized Solution

Geometric Progression as Triangle Sides

The Triangle Inequality

Setting Up the Inequality

Forming the Quadratic Inequality

Finding the Roots

Determining the Valid Range for $r$

Applying the Initial Constraint

The Final Intersection

Evaluating $[r]$

Evaluating $[- r]$

Final Calculation

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

Analyzing the Setup

The Gatekeeper

The Quadratic Challenge

The Final Intersection

Final Calculation

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Enter Numerical Value:

Visualized Solution

Geometric Progression as Triangle Sides

The Triangle Inequality

Setting Up the Inequality

Forming the Quadratic Inequality

Finding the Roots

Determining the Valid Range for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } r

Applying the Initial Constraint

The Final Intersection

Evaluating .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } [r]

Evaluating .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } [−r]

Final Calculation

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

Analyzing the Setup

The Gatekeeper

The Quadratic Challenge

The Final Intersection

Final Calculation

Similar Questions

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Determining the Valid Range for $r$

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