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(B) Let the three numbers in $G.P.$ be $\frac{a}{r}, a, ar$. Since the sequence is increasing,$r > 1$. If the middle number is doubled,the sequence be

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$7+\sqrt{3}$

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(B) Let the three numbers in $G.P.$ be $\frac{a}{r}, a, ar$. Since the sequence is increasing,$r > 1$. If the middle number is doubled,the sequence becomes $\frac{a}{r}, 2a, ar$,which is in $A.P.$ Thus,$2(2a) = \frac{a}{r} + ar$ $\Rightarrow 4 = \frac{1}{r} + r$ $\Rightarrow r^{2} - 4r + 1 = 0$. Solving for $r$,we get $r = \frac{4 \pm \sqrt{16-4}}{2} = 2 \pm \sqrt{3}$. Since the $G.P.$ is increasing,$r = 2 + \sqrt{3}$. The fourth term of the $G.P.$ is $ar^{2} = 3r^{2}$,which implies $a = 3$. The common difference $d$ of the $A.P.$ is $2a - \frac{a}{r} = 2(3) - \frac{3}{2+\sqrt{3}} = 6 - 3(2-\sqrt{3}) = 6 - 6 + 3\sqrt{3} = 3\sqrt{3}$. Now,$r^{2} - d = (2+\sqrt{3})^{2} - 3\sqrt{3} = (4 + 3 + 4\sqrt{3}) - 3\sqrt{3} = 7 + \sqrt{3}$.

Three numbers are in an increasing geometric progression with common ratio $r$. If the middle number is doubled,then the new numbers are in an arithmetic progression with common difference $d$. If the fourth term of the $G.P.$ is $3r^{2}$,then $r^{2}-d$ is equal to:

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