JEE MAIN - Mathematics (2014 (Offline) - No. 15)
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 :
$$2 - \sqrt 3 $$
$$2 + \sqrt 3 $$
$$\sqrt 2 + \sqrt 3 $$
$$3 + \sqrt 2 $$
Explanation
Let $$a,ar,a{r^2}$$ are in $$G.P.$$
According to the question
$$a,2ar,a{r^2}$$ are in $$A.P.$$
$$ \Rightarrow 2 \times 2ar = a + a{r^2}$$
$$ \Rightarrow 4r = 1 + {r^2}$$
$$ \Rightarrow {r^2} - 4r + 1 = 0$$
$$r = {{4 \pm \sqrt {16 - 4} } \over 2} = 2 \pm \sqrt 3 $$
Since $$r > 1$$
$$\therefore$$ $$\pi = 2 - \sqrt 3 $$ is rejected
Hence, $$r = 2 + \sqrt 3 $$
According to the question
$$a,2ar,a{r^2}$$ are in $$A.P.$$
$$ \Rightarrow 2 \times 2ar = a + a{r^2}$$
$$ \Rightarrow 4r = 1 + {r^2}$$
$$ \Rightarrow {r^2} - 4r + 1 = 0$$
$$r = {{4 \pm \sqrt {16 - 4} } \over 2} = 2 \pm \sqrt 3 $$
Since $$r > 1$$
$$\therefore$$ $$\pi = 2 - \sqrt 3 $$ is rejected
Hence, $$r = 2 + \sqrt 3 $$
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