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For x 0, the least value of K, for which 41 + x + 41 - x, K2, 16x + 16-x are three consecutive terms of an A.P., is equal to :
(1) 10
A.P.: 2 K2 = (41 + x + 41 - x ) + (16x + 16-x )
Let t = 4x + 4-x 2 (A.M. G.M.; t = 2 at x = 0)
41 + x + 41 - x = 4t and 16x + 16-x = t2 - 2
K = t2 + 4t - 2, increasing for t 2
K = 4 + 8 - 2 = 10 (at x = 0)
Answer: (1)
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