Let 0

Q: Wait, why do we divide by $xyz$ in the third condition? Isn't that risky if one of them is zero?

That is a fantastic question! In this problem, we are given $0 < z < y < x$, which guarantees that $x, y,$ and $z$ are strictly positive. Since they are non-zero, dividing by their product $xyz$ is perfectly safe and is a standard, powerful technique in competitive algebra to transform products into sums of reciprocals.

Q: Can I solve for $x$ and $z$ individually using a quadratic equation?

You certainly could, but it would be a long and winding road! While you could express $x$ and $z$ as roots of a quadratic equation, the problem asks for $3(x + y + z)^2$. Notice that we only need the sum $(x+z)$ and the value of $y$. By focusing on the symmetric expressions, we save time and avoid unnecessary algebraic complexity.

Q: Is there a constraint I missed? Why is the order $z < y < x$ important?

The order $z < y < x$ ensures that the numbers are distinct and positive, which validates the existence of the progressions. It also helps us confirm that our calculated values for $x, y,$ and $z$ are consistent with the problem statement. Always check your final values against these constraints to ensure your solution is physically and mathematically sound!

Question

Let $0 < z < y < x$ be three real numbers such that $\frac{1}{x}, \frac{1}{y}, \frac{1}{z}$ are in an arithmetic progression and $x, \sqrt{2}y, z$ are in a geometric progression. If $xy + yz + zx = \frac{3}{\sqrt{2}}xyz$, then $3(x + y + z)^2$ is equal to

Sigma Percentile · Accepted Answer

### The Symphony of Sequences

Welcome, fellow traveler on the path to JEE mastery. Today, we are not just solving an algebra problem; we are uncovering the hidden harmony between arithmetic and geometric progressions.

Imagine you are standing before a puzzle where three numbers, $z, y,$ and $x$, are dancing in a specific order: $0 < z < y < x$. They are bound by three distinct rules, and our mission is to find the value of $3(x + y + z)^2$.

### Phase 1: Decoding the Rules

First, we translate the language of sequences into the language of algebra. We are told that the reciprocals $\frac{1}{x}, \frac{1}{y}, \frac{1}{z}$ are in an Arithmetic Progression (A.P.).

The soul of an A.P. is its symmetry: the middle term is the average of its neighbors. Thus, we write:

$$2 	imes \left( \frac{1}{y} ight) = \frac{1}{x} + \frac{1}{z}$$

Let us call this our Equation (1).

Next, we look at the Geometric Progression (G.P.). We are told that $x, \sqrt{2}y, z$ are in G.P.

The core property here is that the square of the middle term equals the product of the extremes. So, $(\sqrt{2}y)^2 = x \cdot z$, which simplifies beautifully to:

$$2y^2 = xz$$

This is our Equation (2).

### Phase 2: The Symmetric Breakthrough

Now, we face the third condition: $xy + yz + zx = \frac{3}{\sqrt{2}}xyz$. At first glance, this looks intimidating.

But here is the secret: whenever you see a symmetric expression involving products of variables, try dividing by the product of all variables. When we divide both sides by $xyz$, the equation transforms into:

$$\frac{xy}{xyz} + \frac{yz}{xyz} + \frac{zx}{xyz} = \frac{3}{\sqrt{2}}$$

This simplifies to:

$$\frac{1}{z} + \frac{1}{x} + \frac{1}{y} = \frac{3}{\sqrt{2}}$$

This is our Equation (3). We have turned a product-heavy equation into a simple sum of reciprocals.

### Phase 3: The Intersection of Logic

Now, the pieces of the puzzle click into place. From Equation (1), we know that $\frac{1}{x} + \frac{1}{z} = \frac{2}{y}$.

Let us substitute this into Equation (3):

$$\frac{2}{y} + \frac{1}{y} = \frac{3}{\sqrt{2}}$$

This simplifies to $\frac{3}{y} = \frac{3}{\sqrt{2}}$, which immediately reveals that $y = \sqrt{2}$. We have found our first anchor point!

With $y = \sqrt{2}$ in our pocket, we return to Equation (2): $xz = 2y^2$. Substituting $y = \sqrt{2}$, we get:

$$xz = 2(\sqrt{2})^2 = 2(2) = 4$$

Now we know the product of $x$ and $z$.

### Phase 4: The Final Calculation

We are almost there. We need the sum $x + y + z$. We already have $y = \sqrt{2}$, so we need $x + z$.

Let us revisit Equation (1) again: $\frac{1}{x} + \frac{1}{z} = \frac{2}{y}$. This is equivalent to:

$$\frac{x+z}{xz} = \frac{2}{y}$$

Substituting our known values, $\frac{x+z}{4} = \frac{2}{\sqrt{2}}$. Since $\frac{2}{\sqrt{2}} = \sqrt{2}$, we have $\frac{x+z}{4} = \sqrt{2}$, which means $x + z = 4\sqrt{2}$.

Finally, the total sum is $x + y + z = (x + z) + y = 4\sqrt{2} + \sqrt{2} = 5\sqrt{2}$. The target expression is $3(x + y + z)^2$.

Substituting our sum, we get:

$$3(5\sqrt{2})^2 = 3(25 	imes 2) = 3(50) = 150$$

And there it is—the elegance of the final result. The final answer is **150**.

Enter Numerical Value:

Visualized Solution

Problem Setup

A.P. Condition

G.P. Condition

Simplifying the Algebraic Equation

Solving for $y$

Finding $x z$

Finding $x + z$

Calculating $x + y + z$

Evaluating the Target Expression

Key Takeaways

The Sigma Insight: Relation Between A.M., G.M., and H.M.

The Symphony of Sequences

Phase 1

Phase 2

Phase 3

Phase 4

Similar Questions

Let $a, b, c$ be positive integers such that $b / a$ is an integer. If $a, b, c$ are in geometric progression and the arithmetic mean of $a, b, c$ is $b + 2$ , then the value of $\frac{a ^{2} + a - 14}{a + 1}$ is \dots.

Let $A_{1}$ and $A_{2}$ be two arithmetic means and $G_{1}, G_{2}$ and $G_{3}$ be three geometric means of two distinct positive numbers. Then $G_{1}^{4} + G_{2}^{4} + G_{3}^{4} + G_{1}^{2} G_{3}^{2}$ is equal to

If $m$ is the A.M. of two distinct real numbers $l$ and $n$ ( $l, n > 1$ ) and $G_{1}, G_{2}$ and $G_{3}$ are three geometric means between $l$ and $n$ , then $G_{1}^{4} + 2 G_{2}^{4} + G_{3}^{4}$ equals:

If $a, b, c$ are in A.P., $a^{2}, b^{2}, c^{2}$ are in H.P., then prove that either $a = b = c$ or $a, b, - c /2$ form a G.P.

Let $x, y > 0$ . If $x^{3} y^{2} = 2^{15}$ , then the least value of $3 x + 2 y$ is

If $a > 0, b > 0$ and $c > 0$ , prove that $(a + b + c) (\frac{1}{a} + \frac{1}{b} + \frac{1}{c}) \geq 9$ .

If the first and the $(2 n - 1)$ st terms of an A.P., a G.P. and an H.P. are equal and their $n$ th terms are $a, b$ and $c$ respectively, then