Let x, y, z be three positive real numbers in a geometric progression such that x < y < z. If 5x, 16y, and 12z are in an arithmetic progression then the common ratio of the geometric progression is
Explanation:
Let the common ratio be ‘k’. ⇒ x = y/k and z = yk
Now considering the AP, 5x = 5y/k, 16y and 12z = 12yk are in arithmetic progression, ∴ 2 × 16y = 5y/k + 12yk ⇒ 32 = 5/k + 12k ⇒ 12k2 – 32k + 5 = 0 ⇒ (2k – 5)(6k – 1) = 0 ⇒ k = 5/2 or 1/6
As x < y < z, k has to be greater than 1. ∴ k = 5/2
Hence, option (b).
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