If loge a, loge b, loge c are in an A.P. and loge a - loge 2b, loge 2b - loge 3c, loge 3c - loge a a

Q: Wait, why can we just drop the logs in the equation $\log_e (b^2) = \log_e (ac)$?

That is a great question! We can do this because the logarithmic function is a one-to-one function. If $\log_e X = \log_e Y$, then $X$ must equal $Y$. It is a fundamental property that allows us to transition from the logarithmic world back into the algebraic world.

Q: Is there a faster way to handle the second A.P. condition? It looks so messy!

It does look intimidating, but the key is to trust the algebra. By grouping the terms and using the property $2y = x + z$, the $\log_e a$ terms cancel out perfectly. This is a classic JEE strategy: don't panic at the sight of complex expressions; often, they are designed to simplify themselves if you just take the first step.

Q: How do I combine the ratios $a:b$ and $b:c$ to get $a:b:c$?

Think of the ratio as a bridge. If $a:b = 3:2$ and $b:c = 3:2$, the common term is $b$. You need to make the value of $b$ identical in both ratios. Once $b$ is the same (in this case, by making it 6), you can simply write the ratio as $a:b:c$.

Question

If $\log_e a, \log_e b, \log_e c$ are in an A.P. and $\log_e a - \log_e 2b, \log_e 2b - \log_e 3c, \log_e 3c - \log_e a$ are also in an A.P, then $a : b : c$ is equal to

Options:

(A) 9 : 6 : 4

(B) 16 : 4 : 1

(C) 25 : 10 : 4

(D) 6 : 3 : 2

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Welcome, fellow traveler, to a problem that beautifully bridges the gap between the logarithmic world and the structured elegance of progressions. When you first look at this problem, you might feel a slight shiver at the sight of logs and A.P. conditions mixed together.

In JEE Advanced, the most complex-looking problems are often just layers of simple, elegant truths waiting to be peeled back. Let us embark on this journey together.

### The First Revelation

We begin with the statement that $\log_e a, \log_e b, \log_e c$ are in an Arithmetic Progression. The definition of an A.P. is our guiding light here. If three terms $x, y, z$ are in A.P., then $2y = x + z$.

Applying this to our logarithmic terms, we get:
$$2 \log_e b = \log_e a + \log_e c$$

Now, let us invoke the magic of logarithms. Using the power rule, $2 \log_e b$ becomes $\log_e (b^2)$. Using the product rule, $\log_e a + \log_e c$ becomes $\log_e (ac)$.

Thus, we arrive at $\log_e (b^2) = \log_e (ac)$. Since the base is the same, we can equate the arguments:
$$b^2 = ac$$

This, my friend, is the hallmark of a Geometric Progression. We have just discovered that $a, b, c$ are in G.P. Keep this result close; it is the anchor for our final answer.

### The Second Condition

Now, we face the second, more intimidating set of terms: $(\log_e a - \log_e 2b), (\log_e 2b - \log_e 3c), (\log_e 3c - \log_e a)$. They are also in A.P.

We apply the same A.P. property:
$$2(\log_e 2b - \log_e 3c) = (\log_e a - \log_e 2b) + (\log_e 3c - \log_e a)$$

Look closely at the right-hand side. We have $\log_e a - \log_e 2b + \log_e 3c - \log_e a$. The $\log_e a$ and $-\log_e a$ terms cancel out!

The right side simplifies to $\log_e 3c - \log_e 2b$. Now, we use the quotient rule: $\log_e x - \log_e y = \log_e (\frac{x}{y})$.

The equation becomes:
$$2 \log_e \left( \frac{2b}{3c} \right) = \log_e \left( \frac{3c}{2b} \right)$$

Using the power rule again, we get $\log_e (\frac{2b}{3c})^2 = \log_e (\frac{3c}{2b})$. Equating the arguments, we have:
$$\left( \frac{2b}{3c} \right)^2 = \frac{3c}{2b}$$

Expanding this, we get $\frac{4b^2}{9c^2} = \frac{3c}{2b}$. Cross-multiplying gives $8b^3 = 27c^3$, which simplifies to $2b = 3c$, or:
$$\frac{b}{c} = \frac{3}{2}$$

### The Synthesis

We are almost there. We know $\frac{b}{c} = \frac{3}{2}$. We also know from our first phase that $b^2 = ac$, which means $\frac{a}{b} = \frac{b}{c}$.

Since $\frac{b}{c} = \frac{3}{2}$, it follows that $\frac{a}{b} = \frac{3}{2}$ as well. Now, we have $a:b = 3:2$ and $b:c = 3:2$.

To combine these into a single ratio $a:b:c$, we must ensure the value of $b$ is the same in both. Multiplying the first ratio by 3 and the second by 2, we get $a:b = 9:6$ and $b:c = 6:4$.

Combining them, we find the final ratio:
$$\mathbf{a:b:c = 9:6:4}$$

We have arrived at the destination. The logic holds, the algebra is sound, and the result is beautiful. Remember, in physics and math, the path is just as important as the answer.

Animated Solution for Mathematics - Sequence and Series: If $lo g_{e} a, lo g_{e} b, lo g_{e} c$ are in an A.P. and $lo g_{e} a - lo g_{e} 2 b, lo g_{e} 2 b - lo g_{e} 3 c, lo g_{e} 3 c - lo g_{e} a$ are also in an A.P, then $a : b : c$ is equal to

Select Answer:

Visualized Solution

First A.P. Condition

Applying A.P. Property

Simplifying with Log Properties

Extracting the Relationship

Analyzing the Second A.P.

Applying A.P. to the Second Set

Simplifying the Right Hand Side

Applying the Quotient Rule

Applying the Power Rule

Equating the Arguments

Solving for the Ratio b/c

Finding the Ratio a/b

Combining the Ratios

Final Ratio a : b : c

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

Analyzing the Setup

The First Revelation

The Second Condition

The Synthesis

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