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

• A. 9 : 6 : 4
• B. 16 : 4 : 1
• C. 25 : 10 : 4
• D. 6 : 3 : 2

Answer 1

The Correct Option is A

To solve this problem, we need to understand the conditions given in the question involving arithmetic progression (A.P.) and logarithms. Let's break it down step by step:

Step 1: Understanding the given sequence in A.P. 

We are given that \(\log_e a, \log_e b, \log_e c\) are in an arithmetic progression. This means that:

• The difference between consecutive terms is constant.
• Thus, we can write: \(\log_e b - \log_e a = \log_e c - \log_e b\)

From the above, it follows that: \(\log_e b - \log_e a = \log_e c - \log_e b = d\) (common difference, \(d\))

Step 2: Given conditions for other terms in A.P.

Another set of terms is also in an A.P.:

• \(\log_e a - \log_e 2b, \log_e 2b - \log_e 3c, \log_e 3c - \log_e a\)

This implies: \((\log_e a - \log_e 2b) - (\log_e 2b - \log_e 3c) = (\log_e 2b - \log_e 3c) - (\log_e 3c - \log_e a)\)

Step 3: Simplifying the expressions

Using properties of logarithms, let's simplify the expressions:

• \(a - b = b - c\)
• Therefore, using exponential forms: \(b^2 = ac\)
• For the second condition, we find: \((\log_e a - \log_e 2b) = \log_e \frac{a}{2b}\)

When resolved: \(\frac{a}{2b} \cdot \frac{2b}{3c} \cdot \frac{3c}{a} = 1\)

Step 4: Solving the equations

Substitute \(b^2 = ac\) into the ratio condition: \(\frac{a}{2b} = \frac{2b}{3c} = \frac{3c}{a}\)

We find that:

• If we assume an identity for ratios like \(k\): \(a = 3k, b = 2k, c = k\)

 

Step 5: Finding the ratio \(a : b : c\)

With the values determined:

• \(a = 3 \times 3 = 9\)
• \(b = 2 \times 3 = 6\)
• \(c = 3 \times 1 = 4\)

 

Thus, the ratio \(a : b : c\) is \(9 : 6 : 4\).

This matches the correct answer given: 9 : 6 : 4.

Answer 2

Since \( \log_e a, \log_e b, \log_e c \) are in an A.P., we have:  
\(b^2 = ac\)

Also, since \( \log_e \left( \frac{a}{2b} \right), \log_e \left( \frac{2b}{3c} \right), \log_e \left( \frac{3c}{a} \right) \) are in an A.P., we get:  
\(\left( \frac{2b}{3c} \right)^2 = \frac{a}{2b} \times \frac{3c}{a}\)

\(\implies \frac{b}{c} = \frac{3}{2}\)
Substituting into equation (1):  
\(b^2 = a \times \frac{2b}{3}\)
\(\implies \frac{a}{b} = \frac{3}{2}\)
Thus, \( a : b : c = 9 : 6 : 4 \).

The Correct Answer is: 9 : 6 : 4