If a, b, c, d and p are distinct real numbers such that (a²+b²+c²)p² - 2(ab+bc+cd)p + (b²+c²+d²) ≤ 0

Q: Wait, why do we assume the sum of squares must be zero? Can't it be negative?

That is a fantastic question! Remember, we are dealing with real numbers $a, b, c, d,$ and $p$. For any real number $X$, $X^2$ is always greater than or equal to zero. When you add three non-negative numbers, their sum can never be negative. Therefore, the only way for the sum to be less than or equal to zero is if the sum is exactly zero.

Q: Why can't we just use the discriminant of the quadratic equation to solve this?

You absolutely could! If you treat the expression as a quadratic in $p$ of the form $Ap^2 + Bp + C \le 0$, you could analyze the discriminant. However, the 'sum of squares' method is often more elegant and faster in competitive exams like JEE, as it directly reveals the relationship between the variables without needing to handle complex inequalities.

Q: How do I know for sure that $a, b, c, d$ are in G.P. from the final equations?

Great observation. We found that $ap-b=0$, $bp-c=0$, and $cp-d=0$. This gives us $p = \frac{b}{a} = \frac{c}{b} = \frac{d}{c}$. Since the ratio of consecutive terms is constant (equal to $p$), this is the definition of a Geometric Progression. You have successfully proven it!

Question

If $a, b, c, d$ and $p$ are distinct real numbers such that $(a^2+b^2+c^2)p^2 - 2(ab+bc+cd)p + (b^2+c^2+d^2) \le 0$ then $a, b, c, d$

Options:

(A) are in A. P.

(B) are in G. P.

(C) are in H. P.

(D) satisfy $ab = cd$

(E) satisfy none of these

Sigma Percentile · Accepted Answer

### The Hidden Elegance of Algebra

Welcome, future engineer! Today, we are going to peel back the layers of a problem that might look like a daunting, messy inequality at first glance.

In the world of JEE Advanced, problems are rarely just about calculation; they are about pattern recognition. Let us look at the given inequality:

$$(a^2+b^2+c^2)p^2 - 2(ab+bc+cd)p + (b^2+c^2+d^2) \le 0$$

At first, this looks like a chaotic jumble of variables. But take a deep breath. Mathematics is the art of finding order in chaos.

### Phase 1: The Art of Expansion

Our first step is to break the expression open. By distributing $p^2$ and $-2p$, we get:

$$a^2p^2 + b^2p^2 + c^2p^2 - 2abp - 2bcp - 2cdp + b^2 + c^2 + d^2 \le 0$$

It looks longer, yes, but now we have individual terms to play with. This is the moment where you must trust the process. We are looking for a structure, specifically the identity $(x-y)^2 = x^2 - 2xy + y^2$.

### Phase 2: The Power of Grouping

Now, let us regroup these terms. We want to pair them up to form perfect squares. Look at the terms involving $a$ and $b$: $a^2p^2 - 2abp + b^2$.

This is a perfect square! It is $(ap-b)^2$. If we continue this logic, we can group the terms involving $b$ and $c$, and then $c$ and $d$.

We get:

$$(a^2p^2 - 2abp + b^2) + (b^2p^2 - 2bcp + c^2) + (c^2p^2 - 2cdp + d^2) \le 0$$

This simplifies beautifully to:

$$(ap-b)^2 + (bp-c)^2 + (cp-d)^2 \le 0$$

### Phase 3: The Logical Leap

Here is the 'Aha!' moment. We have a sum of three squares that is less than or equal to zero. But wait—we know that for any real number $X$, $X^2 \ge 0$.

The sum of three non-negative numbers cannot be negative. Therefore, the only way this inequality holds is if the sum is exactly zero.

This implies that each individual square must be zero:

$$(ap-b)^2 = 0, \quad (bp-c)^2 = 0, \quad (cp-d)^2 = 0$$

### Phase 4: The Geometric Progression

From these equations, we get $ap = b$, $bp = c$, and $cp = d$. This leads us to the ratios:

$$p = \frac{b}{a} = \frac{c}{b} = \frac{d}{c}$$

Because the ratio of any term to its preceding term is a constant $p$, we have confirmed that **$a, b, c, d$ are in a Geometric Progression**.

You have just navigated through a complex algebraic expression to find a beautiful geometric truth. Keep this mindset—always look for the structure, and the answer will reveal itself.

Animated Solution for Mathematics - Sequence and Series: If $a, b, c, d$ and $p$ are distinct real numbers such that $(a^{2} + b^{2} + c^{2}) p^{2} - 2 (ab + b c + c d) p + (b^{2} + c^{2} + d^{2}) \leq 0$ then $a, b, c, d$

Select Answer:

Visualized Solution

Given Inequality for $p$

Expanding the Expression

Regrouping Terms

Forming $(x - y)^{2}$

Property of $X^{2} \geq 0$

Solving for $p$

Condition for G.P.

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

The Hidden Elegance of Algebra

Phase 1

Phase 2

Phase 3

Phase 4

Similar Questions

Let $a, b, c, d$ and $p$ be any non zero distinct real numbers such that $(a^{2} + b^{2} + c^{2}) p^{2} -2 (ab + b c + c d) p + (b^{2} + c^{2} + d^{2}) = 0$ . Then :

If three distinct numbers a,b,c are in G.P. and the equations $a x^{2} + 2 b x + c = 0$ and $d x^{2} + 2 e x + f = 0$ have a common root, then which one of the following statements is correct?

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

If $a_{1} (> 0), a_{2}, a_{3}, a_{4}, a_{5}$ are in a G.P., $a_{2} + a_{4} = 2 a_{3} + 1$ and $3 a_{2} + a_{3} = 2 a_{4}$ , then $a_{2} + a_{4} + 2 a_{5}$ is equal to ______.

If a, b, c are in A.P. and $a^{2}, b^{2}, c^{2}$ are in G.P. such that $a < b < c$ and $a + b + c = \frac{3}{4}$ , then the value of a is :-

In a G.P., if the product of the first three terms is 27 and the set of all possible values for the sum of its first three terms is $R - (a, b)$ , then $a^{2} + b^{2}$ is equal to ......... .

If $x_{1}, x_{2}, x_{3}$ and $y_{1}, y_{2}, y_{3}$ are both in G.P. with the same common ratio, then the points $(x_{1}, y_{1}), (x_{2}, y_{2})$ and $(x_{3}, y_{3})$

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,c,d and p are distinct real numbers such that (a2+b2+c2)p2−2(ab+bc+cd)p+(b2+c2+d2)≤0 then a,b,c,d

Select Answer:

Visualized Solution

Given Inequality for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } p