For three positive integers p, q, r, x^(pq²) = y^(qr) = z^(p²r) and r = pq + 1 such that 3, 3log_y x

Q: Why do we introduce the constant $\lambda$? Can't we just solve it directly?

Great question! While you could try to manipulate the exponents directly, it becomes incredibly messy. Introducing $\lambda$ is a standard, elegant technique in JEE problems involving chains of equalities like $a^x = b^y = c^z$. It acts as a bridge, allowing us to isolate each variable and convert the exponential relationships into simpler logarithmic ratios.

Q: Why does the base change formula simplify things so much?

The base change formula $\log_y x = \frac{\log_\lambda x}{\log_\lambda y}$ is the key to unlocking the problem. By converting all logs to base $\lambda$, we transform the complex logarithmic expressions into simple algebraic fractions involving $p, q,$ and $r$. It turns a logarithmic problem into a system of algebraic equations, which is much easier to solve.

Q: Wait, why do we reject the negative roots or assume $p, q, r$ are positive integers?

The problem explicitly states that $p, q,$ and $r$ are positive integers. This is a crucial constraint! It helps us narrow down the possible values for our variables. When you solve the resulting algebraic equations, you might find multiple solutions, but only the positive integer solutions will satisfy the original conditions of the problem.

Question

For three positive integers $p, q, r$, $x^{pq^2} = y^{qr} = z^{p^2r}$ and $r = pq + 1$ such that $3, 3\log_y x, 3\log_z y, 7\log_x z$ are in A.P. with common difference $\frac{1}{2}$. Then $r - p - q$ is equal to

Options:

(A) 2

(B) 6

(C) 12

(D) -6

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are standing at the base of a mountain, looking up at a complex, interconnected system of equations: $x^{pq^2} = y^{qr} = z^{p^2r}$. In the world of JEE Advanced, complexity is often just a mask for hidden symmetry.

We are given the relationship $r = pq + 1$ and a sequence of terms in an Arithmetic Progression (A.P.): $3, 3\log_y x, 3\log_z y, 7\log_x z$, with a common difference of $d = \frac{1}{2}$. Our mission is to find the value of $r - p - q$.

### Phase 1: The Art of Decoupling

Whenever you see a chain of equalities like $x^{pq^2} = y^{qr} = z^{p^2r}$, your first instinct should be to introduce a common parameter. Let us set this entire chain equal to a constant, $\lambda$.

By doing this, we can write:
$$x^{pq^2} = \lambda, \quad y^{qr} = \lambda, \quad z^{p^2r} = \lambda$$

Taking the logarithm of both sides, we isolate the variables:
$$\log_x \lambda = pq^2, \quad \log_y \lambda = qr, \quad \log_z \lambda = p^2r$$

### Phase 2: The Logarithmic Bridge

To evaluate the terms in our A.P., we utilize the base change formula: $\log_y x = \frac{\log_\lambda x}{\log_\lambda y}$. Since $\log_\lambda x = \frac{1}{pq^2}$ and $\log_\lambda y = \frac{1}{qr}$, we substitute these values:

$$\log_y x = \frac{1/pq^2}{1/qr} = \frac{qr}{pq^2} = \frac{r}{pq}$$

Similarly, we find the other logarithmic ratios:
$$\log_z y = \frac{p^2}{q}, \quad \log_x z = \frac{q^2}{pr}$$

### Phase 3: The Rhythm of the A.P.

The sequence $3, 3\log_y x, 3\log_z y, 7\log_x z$ is an A.P. with $d = \frac{1}{2}$. Calculating the terms:
1. First term: $3$
2. Second term: $3 + \frac{1}{2} = \frac{7}{2}$
3. Third term: $3 + 2(\frac{1}{2}) = 4$
4. Fourth term: $3 + 3(\frac{1}{2}) = \frac{9}{2}$

Equating our derived expressions to these values:
- $3\log_y x = \frac{7}{2} \Rightarrow \log_y x = \frac{7}{6} \Rightarrow \frac{r}{pq} = \frac{7}{6}$
- $3\log_z y = 4 \Rightarrow \log_z y = \frac{4}{3} \Rightarrow \frac{p^2}{q} = \frac{4}{3}$
- $7\log_x z = \frac{9}{2} \Rightarrow \log_x z = \frac{9}{14} \Rightarrow \frac{q^2}{pr} = \frac{9}{14}$

### Phase 4: The Final Unveiling

Using the relation $r = pq + 1$ in the first equation:
$$\frac{pq + 1}{pq} = \frac{7}{6} \Rightarrow 1 + \frac{1}{pq} = \frac{7}{6} \Rightarrow \frac{1}{pq} = \frac{1}{6}$$

Thus, $pq = 6$, which implies $r = 6 + 1 = 7$. Now, substitute $q = \frac{6}{p}$ into the second equation:
$$\frac{p^2}{6/p} = \frac{p^3}{6} = \frac{4}{3} \Rightarrow p^3 = 8 \Rightarrow p = 2$$

With $p = 2$, we find $q = \frac{6}{2} = 3$. The values are $p=2, q=3, r=7$.

The final calculation is:
$$r - p - q = 7 - 2 - 3 = \mathbf{2}$$

Animated Solution for Mathematics - Sequence and Series: For three positive integers $p, q, r$ , $x^{p q^{2}} = y^{q r} = z^{p^{2} r}$ and $r = pq + 1$ such that $3, 3 lo g_{y} x, 3 lo g_{z} y, 7 lo g_{x} z$ are in A.P. with common difference $\frac{1}{2}$ . Then $r - p - q$ is equal to

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Visualized Solution

Given Equations and A.P.

Introducing the Constant $λ$

Logarithmic Conversion

Base Change Formula for $lo g_{y} x$

Calculating $lo g_{z} y$ and $lo g_{x} z$

Evaluating A.P. Terms

Solving for $lo g_{y} x$

Solving for $lo g_{z} y$

Solving for $lo g_{x} z$

Using the Relation $r = pq + 1$

Finding the Value of $r$

Solving for $p$ and $q$

Final Calculation: $r - p - q$

The Sigma Insight: Arithmetic Progression (A.P.)

Analyzing the Setup

Phase 1

Phase 2

Phase 3

Phase 4

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Visualized Solution

Given Equations and A.P.

Introducing the Constant .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } λ

Logarithmic Conversion

Base Change Formula for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } logy​x

Calculating .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } logz​y and .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } logx​z

Evaluating A.P. Terms

Solving for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } logy​x

Solving for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } logz​y

Solving for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } logx​z

Using the Relation .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } r=pq+1

Finding the Value of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } r

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

Final Calculation: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } r−p−q