If a, b and c be three distinct real numbers in G. P. and a + b + c = xb, then x cannot be : (A) 4 (

Q: Why can't we just use the standard $a, ar, ar^2$ notation? It seems more natural.

You absolutely can! However, in competitive exams like JEE Advanced, time and elegance are your best friends. Using $a = \frac{b}{r}$, $b$, and $c = br$ centers the progression around the middle term $b$. When you sum them up, the $b$ factors out beautifully, leaving you with a clean expression involving only $r$ and $x$. It saves you from dealing with quadratic equations that would arise otherwise.

Q: Why do we have to exclude $r = 1$ and $r = -1$?

Great catch! The problem explicitly states that $a, b,$ and $c$ are 'distinct' real numbers. If $r = 1$, then $a = b = c$, which means the terms are identical, not distinct. If $r = -1$, then $a = -b$ and $c = -b$, meaning $a = c$, which again violates the distinctness condition. Always watch out for these subtle constraints in the problem statement; they are often the difference between a correct answer and a trap.

Q: I understand the AM-GM part, but why is the interval $[-1, 3]$ forbidden?

Think of it this way: we found that $x - 1 = r + \frac{1}{r}$. We know that $r + \frac{1}{r}$ can never take values in the open interval $(-2, 2)$. Therefore, $x - 1$ cannot take values in $(-2, 2)$. If we add $1$ to all parts of that inequality, we get $x$ cannot be in $(-2 + 1, 2 + 1)$, which is $(-1, 3)$. Since the boundaries $r=1$ and $r=-1$ are excluded, the forbidden zone is the closed interval $[-1, 3]$.

Question

If a, b and c be three distinct real numbers in G. P. and $a + b + c = xb$, then x cannot be :

Options:

(A) 4

(B) -3

(C) -2

(D) 2

Sigma Percentile · Accepted Answer

### The Symphony of Symmetry: Unlocking the Geometric Progression Welcome, future engineers. Today, we are not just solving an algebra problem; we are peeling back the layers of a mathematical structure. When you see a Geometric Progression (G.P.) in a JEE Advanced problem, your intuition might scream, "Use $a, ar, ar^2$!" While that is technically correct, it is often the path of most resistance. Let us learn to choose the path of elegance. ### Phase 1: The Symmetric Setup Imagine you are standing on a balance beam. If you place your weight at the ends, you wobble, but if you center yourself, you find stability. We apply this same philosophy to our G.P. terms. Instead of the standard notation, let us define our three distinct real numbers as $a = \frac{b}{r}$, $b$, and $c = br$. Why do we do this? Because the problem gives us the sum $a + b + c = xb$. By using this symmetric definition, watch what happens when we substitute these into the equation: $$ \frac{b}{r} + b + br = xb $$ Suddenly, the variable $b$ is everywhere. Since the problem implies a non-trivial progression, we know $b eq 0$. We can divide the entire equation by $b$ with confidence, leaving us with: $$ \frac{1}{r} + 1 + r = x $$ This is the moment of clarity. We have reduced a three-variable problem into a clean relationship between $x$ and the common ratio $r$: $$ r + \frac{1}{r} = x - 1 $$ ### Phase 2: The Forbidden Zone Now, we must analyze the expression $f(r) = r + \frac{1}{r}$. This is a classic function in the JEE syllabus. If you have ever studied the AM-GM inequality, you know that for any positive real number $r$, the sum of a number and its reciprocal is at least $2$. That is, $r + \frac{1}{r} \geq 2$. But what if $r$ is negative? If $r < 0$, let $r = -k$ where $k > 0$. Then $r + \frac{1}{r} = -(k + \frac{1}{k})$. Since $k + \frac{1}{k} \geq 2$, it follows that $r + \frac{1}{r} \leq -2$. This creates a "Forbidden Zone." The value of $r + \frac{1}{r}$ can never exist in the open interval $(-2, 2)$. It is physically impossible for the expression to land there. ### Phase 3: The Constraint of Distinctness We must pause and respect the problem's constraints. We are told the numbers are **distinct**. If $r = 1$, then $r + \frac{1}{r} = 2$. If $r = -1$, then $r + \frac{1}{r} = -2$. Since our terms must be distinct, $r$ cannot be $1$ or $-1$. Consequently, the values $2$ and $-2$ are also forbidden. This upgrades our inequality from "greater than or equal to" to strictly "greater than": $$ \left| r + \frac{1}{r} ight| > 2 $$ ### Phase 4: The Final Verdict We are in the home stretch. We know $r + \frac{1}{r} = x - 1$. Substituting this into our inequality, we get: $$ |x - 1| > 2 $$ This absolute value inequality splits into two distinct paths: 1. $x - 1 > 2 \Rightarrow x > 3$ 2. $x - 1 < -2 \Rightarrow x < -1$ This means $x$ can live anywhere in the range $(-\infty, -1)$ or $(3, \infty)$. Conversely, $x$ is absolutely forbidden from entering the closed interval $[-1, 3]$. When you look at your options, you are testing which value falls into this "forbidden" trap. If you see a value like $2$, you know immediately that it is impossible, because $2$ sits right in the heart of our forbidden interval $[-1, 3]$. Mathematics is not just about finding the answer; it is about understanding the boundaries of what is possible. You have just mapped the territory of this function. Keep this mindset, and no JEE problem will ever be able to hide its secrets from you.

Animated Solution for Mathematics - Sequence and Series: If a, b and c be three distinct real numbers in G. P. and $a + b + c = x b$ , then x cannot be :

Select Answer:

Visualized Solution

Defining the G.P. Terms

The Distinctness Condition

Setting up the Equation

Simplifying the Equation

Isolating $r + \frac{1}{r}$

Graphing $y = r + \frac{1}{r}$

The Standard Bounds

Applying the Distinctness Constraint

The Forbidden Zone

Relating Back to $x$

Solving for $x$ : Case 1

Solving for $x$ : Case 2

The Final Range of $x$

Checking the Options

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

The Symphony of Symmetry

Phase 1

Phase 2

Phase 3

Phase 4

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Animated Solution for Mathematics - Sequence and Series: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If a, b and c be three distinct real numbers in G. P. and a+b+c=xb, then x cannot be :