A) \[\frac{1}{2\sqrt{2}}\]
B) \[\frac{1}{2\sqrt{3}}\]
C) \[\frac{1}{2}-\frac{1}{\sqrt{3}}\]
D) \[\frac{1}{2}-\frac{1}{\sqrt{2}}\]
Correct Answer: D
Solution :
Since,\[a,\,\,b,\,\,c\]are in\[AP\]. \[a=A-D,\,\,b=A,\,\,c=A+D\] Where,\[A\]is the first term and\[D\]is the common difference of an\[AP\]. Given, \[a+b+c=\frac{3}{2}\] \[\Rightarrow \] \[(A-D)+A+(A+D)=\frac{3}{2}\] \[\Rightarrow \] \[3A=\frac{3}{2}\] \[\Rightarrow \] \[A=\frac{1}{2}\] \[\therefore \]The numbers are\[\frac{1}{2}-D,\,\,\frac{1}{2},\,\,\frac{1}{2}+D\] Also,\[{{\left( \frac{1}{2}-D \right)}^{2}},\,\,\frac{1}{4},\,\,{{\left( \frac{1}{2}+D \right)}^{2}}\]are in\[GP\] \[\Rightarrow \] \[{{\left( \frac{1}{4} \right)}^{2}}={{\left( \frac{1}{2}-D \right)}^{2}}{{\left( \frac{1}{2}+D \right)}^{2}}\] \[\Rightarrow \] \[\frac{1}{16}={{\left( \frac{1}{4}-{{D}^{2}} \right)}^{2}}\] \[\Rightarrow \] \[\frac{1}{4}-{{D}^{2}}=\pm \frac{1}{4}\] \[\Rightarrow \] \[{{D}^{2}}=\frac{1}{2}\] (\[\because \,\,D=0\]is not possible) \[\Rightarrow \] \[D=\pm \frac{1}{\sqrt{2}}\] \[\therefore \] \[a=\frac{1}{2}\pm \frac{1}{\sqrt{2}}\] So, out of the given value\[a=\frac{1}{2}-\frac{1}{\sqrt{2}}\]is the right choice.
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