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
- 9 : 6 : 4
- 16 : 4 : 1
- 25 : 10 : 4
- 6 : 3 : 2
The Correct Option is A
Approach Solution - 1
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.
Approach Solution -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
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