A) \[1-\sqrt{3}:2:1+\sqrt{3}\]
B) \[1:\sqrt{2}:-\sqrt{3}\]
C) \[1:-\sqrt{2}:\sqrt{3}\]
D) \[1\mp \sqrt{3}:-2:1\pm \sqrt{3}\]
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. \[\therefore \]\[2{{q}^{2}}={{p}^{2}}+{{r}^{2}}={{(p+r)}^{2}}-2pr\] \[\Rightarrow \] \[8{{K}^{2}}={{(p+r)}^{2}}-2(p+r)K\] \[\Rightarrow \]\[{{(p+r)}^{2}}-2(p+r)K-8{{K}^{2}}=0\]\[\Rightarrow \]\[p+r=4K,\ -2K\] When\[p+r=4K\], then \[pr=4{{K}^{2}}\] \[\therefore \]\[{{(p-r)}^{2}}={{(p+r)}^{2}}-4pr=16{{K}^{2}}-16{{K}^{2}}=0\]\[\Rightarrow \]\[p=r\] But this is not possible \[(\because \ \ p\ne r)\] \ \[p+r=-2K\]\[\Rightarrow \]\[pr=-2K\ .\ K=-2{{K}^{2}}\] Now \[{{(p-r)}^{2}}={{(p+r)}^{2}}-4pr=4{{K}^{2}}-4(-2{{K}^{2}})=12{{K}^{2}}\] \[\Rightarrow \]\[p-r=\pm 2\sqrt{3}K,\] also \[p+r=-2K\] \[\therefore \]\[2p=(-2\pm 2\sqrt{3})K\]\[\Rightarrow \]\[p=(-1\pm \sqrt{3})K\] and \[2r=-2K\mp 2\sqrt{3}K\]\[\Rightarrow \]\[r=(-1\mp \sqrt{3})K\] \[\therefore \]\[p:q:r=(-1\pm \sqrt{3})K:2K:(-1\mp \sqrt{3})K\] \[=-1\pm \sqrt{3}:2:-1\mp \sqrt{3}=1\mp \sqrt{3}:-2:1\pm \sqrt{3}\].
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