If for x, y ∈ R, x 0, y = log₁₀ x + log₁₀ x¹/³ + log₁₀ x¹/⁹ + .... upto ∞ terms and (2+4+6+....+2y

Q: Wait, why can we just cancel the (1+2+3+\dots+y) term? Is it possible that it equals zero?

That is a sharp observation! In mathematics, we must always be careful about division by zero. However, look at the context: x > 0. The series 1+2+3+\dots+y represents a sum of positive integers. Since y must be a positive value for the series to exist in this context, the sum is strictly positive and non-zero. Therefore, cancelling it is perfectly valid and mathematically sound.

Q: I tried to solve the infinite series first, but I got stuck because I didn't know x. Is there a specific order to these problems?

You have stumbled upon a classic JEE strategy! When you see a system of equations, always scan for the 'key'—the equation that provides a concrete value. The infinite series for y depends on x, but the fraction equation allows us to isolate \log_{10} x immediately. Always solve for the independent variable first to unlock the rest of the problem.

Q: Why is the sum of the GP \frac{3}{2}? I thought the formula was \frac{a}{1-r}?

You are absolutely correct about the formula! Let's check the values: the first term a is 1, and the common ratio r is \frac{1}{3}. Plugging these into \frac{a}{1-r}, we get \frac{1}{1 - 1/3}. The denominator becomes \frac{2}{3}. Dividing 1 by \frac{2}{3} is the same as multiplying by its reciprocal, which is \frac{3}{2}. Your intuition is spot on—keep trusting your formulas!

Question

If for $x, y \in \mathbb{R}, x > 0$\ $y = \log_{10} x + \log_{10} x^{1/3} + \log_{10} x^{1/9} + \dots$ upto $\infty$ terms\ and $\frac{2+4+6+\dots+2y}{3+6+9+\dots+3y} = \frac{4}{\log_{10} x}$, then the ordered pair $(x, y)$ is equal to :

Options:

(A) $(10^6, 6)$

(B) $(10^4, 6)$

(C) $(10^2, 3)$

(D) $(10^6, 9)$

Sigma Percentile · Accepted Answer

### The Illusion of Complexity

Look at the second equation provided:

$$\frac{2+4+6+\dots+2y}{3+6+9+\dots+3y} = \frac{4}{\log_{10} x}$$

Your eyes might dart to the numerator and denominator, seeing an arithmetic progression that goes on until $y$. You might be tempted to use the sum formula for an arithmetic progression, $\frac{n}{2}(2a + (n-1)d)$.

But stop. Before you dive into heavy algebra, look for the symmetry. In the numerator, every term is a multiple of $2$. In the denominator, every term is a multiple of $3$.

If we factor out a $2$ from the top and a $3$ from the bottom, we are left with:

$$\frac{2(1+2+3+\dots+y)}{3(1+2+3+\dots+y)} = \frac{4}{\log_{10} x}$$

Suddenly, the monster shrinks. The term $(1+2+3+\dots+y)$ is common to both the numerator and the denominator. Since we know $x > 0$, the logarithm is defined, and the sum is non-zero, we can cancel it out with confidence.

We are left with the beautifully simple ratio:

$$\frac{2}{3} = \frac{4}{\log_{10} x}$$

This is the turning point. By cross-multiplying, we find that $\log_{10} x = \frac{4 	imes 3}{2}$, which simplifies to $6$. Just like that, the first piece of our puzzle, $x$, is revealed as $x = 10^6$.

### The Infinite Symphony

Now that we have unlocked $x$, let us turn our attention to the first equation:

$$y = \log_{10} x + \log_{10} x^{\frac{1}{3}} + \log_{10} x^{\frac{1}{9}} + \dots$$

This looks like an infinite series, and infinity can be scary. But remember the power rule of logarithms: $\log_{b}(a^n) = n \log_{b}(a)$.

This rule is your best friend. It allows us to pull those exponents down from their perch. The series becomes:

$$y = \log_{10} x + \frac{1}{3}\log_{10} x + \frac{1}{9}\log_{10} x + \dots$$

Notice that every single term now contains a common factor of $\log_{10} x$. We can factor it out, leaving us with:

$$y = (\log_{10} x) 	imes \left(1 + \frac{1}{3} + \frac{1}{9} + \dotsight)$$

Now, look inside those parentheses. $1, 1/3, 1/9 \dots$ is a classic infinite Geometric Progression (G.P.) where the first term $a = 1$ and the common ratio $r = 1/3$. Since the absolute value of $r$ is less than $1$, the sum converges.

The formula for the sum of an infinite G.P. is $S_{\infty} = \frac{a}{1-r}$. Substituting our values, we get:

$$S_{\infty} = \frac{1}{1 - 1/3} = \frac{1}{2/3} = \frac{3}{2}$$

### The Final Synthesis

We are almost at the finish line. We have the value of the infinite series, which is $3/2$, and we have the value of $\log_{10} x$, which is $6$.

Putting them together, we get:

$$y = 6 	imes \frac{3}{2}$$

Calculating this, $6$ divided by $2$ is $3$, and $3$ times $3$ is $9$. So, $y = 9$.

We have navigated the maze. We found $x = 10^6$ and $y = 9$. The ordered pair $(x, y)$ is **$(10^6, 9)$**.

This problem wasn't about brute force; it was about pattern recognition and the courage to simplify. Never let the notation scare you. Break it down, find the symmetry, and trust your tools.

Select Answer:

Visualized Solution

Analyze the Given Equations

Factorizing the Fraction

Simplifying the Series

Solving for $lo g_{10} x$

Calculating $x$

Analyzing the Infinite Series

Applying the Power Rule

Sum of Infinite G.P.

Evaluating the G.P. Sum

Calculating $y$

Final Answer

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

The Illusion of Complexity

The Infinite Symphony

The Final Synthesis

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