• # question_answer  Suppose $a,\,b,\,c$ are in A.P. and ${{a}^{2}},{{b}^{2}},{{c}^{2}}$ are in G.P. If       a < b < c  and $a+b+c=\frac{3}{2}$, then the value of a is [IIT Screening 2002] A) $\frac{1}{2\sqrt{2}}$ B) $\frac{1}{2\sqrt{3}}$ C) $\frac{1}{2}-\frac{1}{\sqrt{3}}$ D) $\frac{1}{2}-\frac{1}{\sqrt{2}}$

$b=a+d,c=a+2d$, 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$a=\frac{1}{2}-\frac{1}{\sqrt{2}}$.