Let p=\ln(x), q=\ln(x^{3}) and r=\ln(x^{5}), where x \gt 1. Which of the following statements is/are correct? I. p, q and r are in AP. II. p, q and r can <strong>NEVER</strong> be in GP. Select the answer using the code given below.
- A. I only
- B. II only
- C. Both I and II ✓
- D. Neither I nor II
Correct Answer: C. Both I and II
Explanation
We have p = \ln x, q = 3\ln x, and r = 5\ln x. The common difference is 2\ln x, so they are in AP. For them to be in GP, we would need q^2 = pr, meaning 9(\ln x)^2 = 5(\ln x)^2, which is impossible since x \gt 1 implies \ln x \neq 0.
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