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Q.
Three positive numbers form an increasing G.P. If the middle term in this G.P. is doubled, the new numbers are in A.P. Then the common ratio of the G.P. is
Sequences and Series
Solution:
Three+ve numbers
$ a,ar, ar^{2}$ are in $G.P. \quad\left[r>1\right] $
$ \therefore a, 2ar, ar^{2}$ are in $ A.P.$ [ $\because G.P.{\text{ is increasing}}] $
$\Rightarrow 2\left(2ar\right) = a+ar^{2} $
$ \Rightarrow 4r= 1+r^{2} $
$ \Rightarrow r^{2} -4r+1 = 0 $
$ \Rightarrow r= \frac{4\pm\sqrt{16-4}}{2} =\frac{ 4\pm\sqrt{12}}{2} $
$= 2\pm\sqrt{3} $
Since $r > 1$
$ \therefore r^{2}= 2+\sqrt{3}$