Let p, q, r be nonzero real numbers that are, respectively, the 10ᵗʰ, 100ᵗʰ and 1000ᵗʰ terms of a ha

Q: Wait, why do we divide by $pqr$? Is that just a random guess?

It is not a guess, but a strategic move! In systems of equations where coefficients are products of variables (like $qr, pr, pq$), dividing by the product $pqr$ is a standard, powerful technique. It transforms a product-heavy equation into a sum of reciprocals, which aligns perfectly with the properties of Harmonic Progressions. Always look for this symmetry when you see cyclic coefficients!

Q: Why does $A=D$ lead to infinitely many solutions?

Great question. When $A=D$, the third equation becomes a linear combination of the first two. Geometrically, this means the third plane is not independent; it passes through the line of intersection formed by the first two planes. Since the third plane 'contains' the intersection line of the other two, any point on that line satisfies all three equations, leading to infinitely many solutions.

Q: What happens if $A \neq D$? Why is there 'no solution'?

If $A \neq D$, the normal vector of the third plane is still coplanar with the normals of the first two, but the third plane itself is shifted. It does not pass through the intersection line of the first two planes. Instead, the three planes form a triangular prism. Since there is no single point common to all three planes, the system is inconsistent, meaning there is no solution.

Question

Let $p, q, r$ be nonzero real numbers that are, respectively, the $10^{th}, 100^{th}$ and $1000^{th}$ terms of a harmonic progression. Consider the system of linear equations
$x + y + z = 1$
$10x + 100y + 1000z = 0$
$qr x + pr y + pq z = 0$.

List-I:

P. If $q/r = 10$, then the system of linear equations has

Q. If $p/r 
e 100$, then the system of linear equations has

R. If $p/q 
e 10$, then the system of linear equations has

S. If $p/q = 10$, then the system of linear equations has

List-II:

1. $x = 0, y = 10/9, z = -1/9$ as a solution

2. $x = 10/9, y = -1/9, z = 0$ as a solution

3. infinitely many solutions

4. no solution

5. at least one solution

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Welcome, student. Today, we are going to peel back the layers of a problem that, at first glance, looks like a daunting algebraic mess. We have a system of three linear equations, and the coefficients are tied to a Harmonic Progression.

It feels like we are juggling too many variables—$p, q, r, x, y, z$—but I want you to take a deep breath. In JEE Advanced, the complexity is often just a veil. Our job is to lift that veil and find the elegant simplicity underneath.

### Phase 1: The Algebraic Transformation

Let us look at our system:

1. $x + y + z = 1$

2. $10x + 100y + 1000z = 0$

3. $qr x + pr y + pq z = 0$

The first two equations are standard planes in 3D space. But the third equation is the key to the entire puzzle. It is symmetric, yet cluttered.

Whenever you see coefficients like $qr, pr, pq$, your mathematical intuition should scream, "Divide by $pqr$!"

Why? Because when we divide the third equation by $pqr$, we get:

$$\frac{qr x}{pqr} + \frac{pr y}{pqr} + \frac{pq z}{pqr} = 0$$

Which simplifies to the remarkably clean:

$$\frac{x}{p} + \frac{y}{q} + \frac{z}{r} = 0$$

Suddenly, the variables $p, q, r$ are no longer multiplying our unknowns; they are acting as denominators. This is the bridge we needed to cross.

### Phase 2: The Harmonic Bridge

The problem tells us that $p, q, r$ are the $10^{th}, 100^{th},$ and $1000^{th}$ terms of a Harmonic Progression (H.P.). By definition, the reciprocals of terms in an H.P. form an Arithmetic Progression (A.P.).

Let us define this A.P. with a first term $A$ and a common difference $D$. The general term is $T_n = A + (n-1)D$. Therefore, we can write:

$$\frac{1}{p} = A + 9D$$

$$\frac{1}{q} = A + 99D$$

$$\frac{1}{r} = A + 999D$$

This is the heart of the problem. We have translated the abstract concept of an H.P. into the concrete, manageable language of an A.P. Now, let us substitute these back into our simplified third equation:

$$x(A + 9D) + y(A + 99D) + z(A + 999D) = 0$$

### Phase 3: The Moment of Truth

Now, let us group the terms. This is where the magic happens. We collect everything associated with $A$ and everything associated with $D$:

$$A(x + y + z) + D(9x + 99y + 999z) = 0$$

Look at that! We have $(x + y + z)$ and $(9x + 99y + 999z)$. We know from our first equation that $x + y + z = 1$.

But what about the second part? Let us look at our first two equations again:

$$x + y + z = 1$$

$$10x + 100y + 1000z = 0$$

If we subtract the first from the second, we get:

$$(10x - x) + (100y - y) + (1000z - z) = 0 - 1$$

$$9x + 99y + 999z = -1$$

Do you see it? The coefficient of $D$ is exactly $-1$. Our equation collapses into:

$$A(1) + D(-1) = 0 \Rightarrow A = D$$

This is the condition for consistency. If **$A = D$**, the system is consistent. If **$A 
eq D$**, the system breaks down.

### Phase 4: The Geometric Grand Finale

Let us visualize what is happening in 3D space. Each equation represents a plane. When $A = D$, the normal vector of the third plane becomes a linear combination of the first two.

Specifically, the third plane becomes redundant—it is essentially the same as the second plane. We are left with two intersecting planes, which meet along a line. This is why we get **infinitely many solutions**.

Conversely, if $A 
eq D$, the third plane is parallel to the intersection line of the first two but does not contain it. The three planes form a triangular prism.

Imagine three walls of a room that never meet at a single corner. That is why we get **no solution**. This problem is a beautiful reminder that algebra and geometry are two sides of the same coin.

List-I

List-II

Select Matching Pairs:

Visualized Solution

The System of Equations

Simplifying the Third Equation

Harmonic Progression Setup

Expressing the Reciprocals

Substituting into Plane 3

Grouping Variables

Extracting the $D$ -term

The Consistency Condition

Geometric Interpretation

The Infinite Solutions Case

The Inconsistent Case

The Sigma Insight: Solution of System of Linear Equations (Matrix Method and Cramer's Rule)

Analyzing the Setup

Phase 1

Phase 2

Phase 3

Phase 4

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