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 :
\[b=a+d,c=a+2d\], where \[d>0\] Now \[{{a}^{2}},{{(a+d)}^{2}},{{(a+2d)}^{2}}\]are in G.P. \[\therefore {{(a+d)}^{4}}={{a}^{2}}{{(a+2d)}^{2}}\] or \[{{(a+d)}^{2}}=\pm a(a+2d)\] or \[{{a}^{2}}+{{d}^{2}}+2ad=\pm \,\,({{a}^{2}}+2ad)\] Taking (+) sign, d = 0 (not possible as \[a<b<c)\] Taking (-) sign, \[2{{a}^{2}}+4ad+{{d}^{2}}=0\], \[\left[ \because a+b+c=\frac{3}{2},\,\,\,\therefore a+d=\frac{1}{2} \right]\] \[2{{a}^{2}}+4a\left( \frac{1}{2}-a \right)+{{\left( \frac{1}{2}-a \right)}^{2}}=0\] or \[4{{a}^{2}}-4a-1=0\] \[\therefore a=\frac{1}{2}\pm \frac{1}{\sqrt{2}}.\text{Here}\,\text{ }d=\frac{1}{2}-a>0.\]So, \[a<\frac{1}{2}.\] Hence\[a=\frac{1}{2}-\frac{1}{\sqrt{2}}\].
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