If a, b, c are in AP; b, c, d are in GP; c, d, e are in HP, then which of the following is/are correct? 1. a, c and e are in GP 2. \frac{1}{a}, \frac{1}{c}, \frac{1}{e} are in GP Select the correct answer using the code given below:
- A. 1 only
- B. 2 only
- C. Both 1 and 2 ✓
- D. Neither 1 nor 2
Correct Answer: C. Both 1 and 2
Explanation
Given 2b = a+c, c^2 = bd, and d = \frac{2ce}{c+e}. Substitute b and d into the GP condition: c^2 = \left(\frac{a+c}{2}\right) \left(\frac{2ce}{c+e}\right). Solving yields c^2(c+e) = (a+c)ce \implies c^2 = ae. Thus, a, c, e are in GP. Consequently, their reciprocals \frac{1}{a}, \frac{1}{c}, \frac{1}{e} are also in GP.
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