I. \(p, q\) and \(r\) are in AP.
II. \(p, q\) and \(r\) can never be in GP.
Select the answer using the code given below.
Understanding Logarithm Properties and Sequences
The question asks us to analyze three logarithmic expressions, \(p = \ln(x)\), \(q = \ln(x^3)\), and \(r = \ln(x^5)\), where \(x > 1\). We need to determine if these expressions form an Arithmetic Progression (AP) and if they can ever form a Geometric Progression (GP).
Simplifying the Logarithmic Expressions
First, let's simplify the expressions for \(q\) and \(r\) using the power rule of logarithms, which states that \(\ln(a^b) = b \ln(a)\).
- \(p = \ln(x)\)
- \(q = \ln(x^3) = 3 \ln(x)\)
- \(r = \ln(x^5) = 5 \ln(x)\)
Let \(y = \ln(x)\). Since \(x > 1\), we know that \(y > 0\). The expressions become:
- \(p = y\)
- \(q = 3y\)
- \(r = 5y\)
Analysis of Statement I: Arithmetic Progression (AP)
For a sequence of three terms \(a, b, c\) to be in AP, the difference between consecutive terms must be constant. That is, \(b - a = c - b\).
Let's check this condition for \(p, q, r\):
- Calculate the difference between \(q\) and \(p\): \(q - p = 3y - y = 2y\)
- Calculate the difference between \(r\) and \(q\): \(r - q = 5y - 3y = 2y\)
Since \(q - p = 2y\) and \(r - q = 2y\), the difference is constant (\(2y\)). Therefore, the terms \(p, q, r\) are always in an Arithmetic Progression for any \(x > 1\).
Conclusion for Statement I: Statement I is correct.
Analysis of Statement II: Geometric Progression (GP)
For a sequence of three terms \(a, b, c\) to be in GP, the ratio between consecutive terms must be constant. That is, \(b/a = c/b\), provided \(a\) and \(b\) are not zero.
Let's check this condition for \(p, q, r\). We know \(p = y\), \(q = 3y\), and \(r = 5y\). Since \(x > 1\), \(y = \ln(x) > 0\). Thus, \(p, q, r\) are all non-zero.
- Calculate the ratio between \(q\) and \(p\): \(q/p = (3y) / y = 3\)
- Calculate the ratio between \(r\) and \(q\): \(r/q = (5y) / (3y) = 5/3\)
The ratio \(q/p = 3\) is not equal to the ratio \(r/q = 5/3\). Since the common ratio is not constant, the terms \(p, q, r\) are never in a Geometric Progression for \(x > 1\).
Conclusion for Statement II: The statement "p, q and r can never be in GP" is correct because we have shown they are not in GP for any valid \(x\).
Final Conclusion
Both Statement I and Statement II have been found to be correct.
Therefore, the correct option is the one that states both I and II are correct.
Similar Questions
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Important Questions from Logarithms
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