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Step-by-Step Solution
Step 1: Represent the three terms of the G.P.
Let the three positive numbers that form an increasing Geometric Progression (G.P.) be:
$a,\; a r,\; a r^2$, where $a > 0$ and $r > 1$.
Step 2: Form the new sequence by doubling the middle term
The problem states that if the middle term of the G.P. is doubled, the resulting sequence is an Arithmetic Progression (A.P.). Hence, the new terms are:
$a,\; 2ar,\; a r^2.$
Step 3: Apply the property of an A.P.
Three numbers $x,\, y,\, z$ are in A.P. if $2y = x + z.$ Here,
$x = a,\, y = 2ar,\, z = a r^2.$
So, the condition for A.P. becomes:
$$
2 \times (2ar) = a + a r^2.
$$
Simplifying, we get:
$$
4ar = a + ar^2.
$$
Step 4: Simplify and form the quadratic equation in $r$
Divide both sides of $4ar = a + ar^2$ by $a$ (since $a \neq 0$):
$$
4r = 1 + r^2.
$$
Rearranging,
$$
r^2 - 4r + 1 = 0.
$$
Step 5: Solve the quadratic equation
Use the quadratic formula $r = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ for the equation $r^2 - 4r + 1 = 0,$ where $a=1,\; b=-4,\; c=1.$
$$
r = \frac{4 \pm \sqrt{16 - 4}}{2} = \frac{4 \pm \sqrt{12}}{2} = 2 \pm \sqrt{3}.
$$
Step 6: Choose the valid value of $r$
Since the numbers form an increasing G.P., we need $r > 1.$ Of the two solutions $2 \pm \sqrt{3},$ only $2 + \sqrt{3}$ is greater than 1. Hence,
$$
r = 2 + \sqrt{3}.
$$
Final Answer
The common ratio of the G.P. is $2 + \sqrt{3}.$
