Sequence And Series Question 261
Question: If three unequal numbers $ p,\ q,\ r $ are in H.P. and their squares are in A.P., then the ratio $ p:q:r $ is
Options:
A) $ 1-\sqrt{3}:2:1+\sqrt{3} $
B) $ 1:\sqrt{2}:-\sqrt{3} $
C) $ 1:-\sqrt{2}:\sqrt{3} $
D) $ 1\mp \sqrt{3}:-2:1\pm \sqrt{3} $
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Answer:
Correct Answer: D
Solution:
By hypothesis, $ q=\frac{2pr}{p+r} $
$ \Rightarrow $ $ \frac{q}{2}=\frac{pr}{p+r}=K $ (say)
$ \Rightarrow $ $ q=2K,\ pr=(p+r)K. $ Also $ p^{2},\ q^{2},\ r^{2} $ are in A.P.
$ \therefore $ $ 2q^{2}=p^{2}+r^{2}={{(p+r)}^{2}}-2pr $
$ \Rightarrow $ $ 8K^{2}={{(p+r)}^{2}}-2(p+r)K $
$ \Rightarrow $ $ {{(p+r)}^{2}}-2(p+r)K-8K^{2}=0 $
$ \Rightarrow $ $ p+r=4K,\ -2K $ When $ p+r=4K $ , then $ pr=4K^{2} $
$ \therefore $ $ {{(p-r)}^{2}}={{(p+r)}^{2}}-4pr=16K^{2}-16K^{2}=0 $
$ \Rightarrow $ $ p=r $ But this is not possible $ (\because \ \ p\ne r) $ \ $ p+r=-2K $
$ \Rightarrow $ $ pr=-2K\ .\ K=-2K^{2} $ Now $ {{(p-r)}^{2}}={{(p+r)}^{2}}-4pr=4K^{2}-4(-2K^{2})=12K^{2} $
$ \Rightarrow $ $ p-r=\pm 2\sqrt{3}K, $ also $ p+r=-2K $
$ \therefore $ $ 2p=(-2\pm 2\sqrt{3})K $
$ \Rightarrow $ $ p=(-1\pm \sqrt{3})K $ and $ 2r=-2K\mp 2\sqrt{3}K $
$ \Rightarrow $ $ r=(-1\mp \sqrt{3})K $
$ \therefore $ $ p:q:r=(-1\pm \sqrt{3})K:2K:(-1\mp \sqrt{3})K $ $ =-1\pm \sqrt{3}:2:-1\mp \sqrt{3}=1\mp \sqrt{3}:-2:1\pm \sqrt{3} $ .
BETA
