Three positive integers form an increasing GP. If the middle term in this GP is doubled, then new numbers are in AP. Then the common ratio of GP is

Solution :

Let a, ar, \(ar^2\) are in GP (r > 1)

According to the question, a, 2ar, \(ar^2\) in AP.

\(\implies\)  4ar = a + \(ar^2\)

\(\implies\) \(r^2\) – 4r + 1 = 0

\(\implies\) r = \(2 \pm \sqrt{3}\)

Hence, r = \(2 + \sqrt{3}\)    [ \(\because\)  AP is increasing]

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