# 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 &gt; 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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