If x, y and z are pth, qth and rth terms respectively of an A.P. and also of a G.P., then x^(y-z) y^

Q: Wait, why do we reject the negative root here? Or can we just use energy conservation?

That is a great question, but remember that this is an algebraic problem, not a physics one! We are dealing with properties of progressions. There is no 'negative root' to reject here because we are not solving a quadratic equation. We are simply manipulating algebraic expressions using the definitions of A.P. and G.P. Stick to the definitions, and you will see the symmetry emerge naturally.

Q: Why can't I just pick specific values for $p, q, r$ and $A, D, a, R$ to solve this?

You absolutely can! In a competitive exam like JEE, if you are stuck, picking simple values (like $p=1, q=2, r=3$) is a brilliant strategy to check your answer. However, the algebraic proof is more robust and helps you understand the underlying structure. The beauty of this problem lies in the fact that it holds true for any $p, q, r$ and any $A, D, a, R$.

Q: What if $R=1$ or $D=0$? Does the formula still work?

Excellent observation! If $D=0$, the A.P. is constant, and if $R=1$, the G.P. is constant. In both cases, the expression becomes $x^0 \cdot y^0 \cdot z^0$, which is $1 \cdot 1 \cdot 1 = 1$. The formula holds perfectly. It is a beautiful, universal result.

Question

If $x, y$ and $z$ are $p$th, $q$th and $r$th terms respectively of an A.P. and also of a G.P., then $x^{y-z} y^{z-x} z^{x-y}$ is equal to :

Options:

(A) xyz

(B) 0

(C) 1

(D) None of these

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Welcome, fellow traveler on the path to JEE excellence. Today, we are going to dismantle a problem that, at first glance, looks like a chaotic mess of variables.

We are given that $x, y,$ and $z$ are the $p^{th}, q^{th},$ and $r^{th}$ terms of both an Arithmetic Progression (A.P.) and a Geometric Progression (G.P.). Our mission is to evaluate the expression $x^{y-z} y^{z-x} z^{x-y}$.

In the world of JEE, such intimidating expressions are often hiding a secret, a beautiful symmetry waiting to be revealed. Let us embark on this journey together.

### Decoding the Arithmetic Progression

First, let us focus on the A.P. Let the first term be $A$ and the common difference be $D$. The $n^{th}$ term of an A.P. is given by $T_n = A + (n-1)D$.

Therefore, we can write our variables as:
$$x = A + (p-1)D$$
$$y = A + (q-1)D$$
$$z = A + (r-1)D$$

Now, look at the exponents in our target expression: $(y-z), (z-x),$ and $(x-y)$. If we subtract these equations, the first term $A$ cancels out completely.

We are left with:
$$y - z = (q - r)D$$
$$z - x = (r - p)D$$
$$x - y = (p - q)D$$

### Decoding the Geometric Progression

Now, let us turn to the G.P. Let the first term be $a$ and the common ratio be $R$. The $n^{th}$ term of a G.P. is $T_n = aR^{n-1}$.

Thus, we have:
$$x = aR^{p-1}$$
$$y = aR^{q-1}$$
$$z = aR^{r-1}$$

We have now defined our bases ($x, y, z$) and our exponents ($y-z, z-x, x-y$) in terms of the fundamental parameters of the progressions. It is time for the grand synthesis.

### The Algebraic Symphony

We substitute our G.P. expressions for the bases and our A.P. expressions for the exponents into the target expression:
$$x^{y-z} y^{z-x} z^{x-y} = (aR^{p-1})^{(q-r)D} \cdot (aR^{q-1})^{(r-p)D} \cdot (aR^{r-1})^{(p-q)D}$$

Let us separate the base $a$ and the base $R$. For the base $a$, we have $a$ raised to the power of:
$$D(q-r) + D(r-p) + D(p-q) = D(q-r+r-p+p-q)$$

Inside the parenthesis, all terms cancel out to zero. Thus, $a^0 = 1$.

Now, for the base $R$, the exponent is:
$$D[(p-1)(q-r) + (q-1)(r-p) + (r-1)(p-q)]$$

Expanding this carefully:
$$(pq - pr - q + r) + (qr - qp - r + p) + (rp - rq - p + q)$$

Every term has an exact additive inverse. The entire sum is zero, resulting in $R^0 = 1$.

### Final Calculation

The entire expression collapses to:
$$1 \cdot 1 = 1$$

The complexity was merely an illusion. The final result is **1**.

Animated Solution for Mathematics - Sequence and Series: If $x, y$ and $z$ are $p$ th, $q$ th and $r$ th terms respectively of an A.P. and also of a G.P., then $x^{y - z} y^{z - x} z^{x - y}$ is equal to :

Select Answer:

Visualized Solution

Understanding the Problem

Defining A.P. Terms

Calculating the Exponents

Defining G.P. Terms

Substitution into Expression

Simplifying Base $a$

Simplifying Base $R$

Final Result

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

Analyzing the Setup

Decoding the Arithmetic Progression

Decoding the Geometric Progression

The Algebraic Symphony

Final Calculation

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