Question

If $a,b,c$ are in A.P. and $a^2,b^2,c^2$ are in G.P. such that $a<b<c$ and $a+b+c=\frac{3}{4}$, then the value of $a$ is :

• A. $\frac{1}{4}$
• B. $\frac{1}{4}-\frac{1}{2\sqrt{2}}$
• C. $\frac{1}{4}+\frac{1}{2\sqrt{2}}$
• D. $\frac{1}{4}-\frac{1}{4\sqrt{2}}$

Answer 1

The correct option is B $\frac{1}{4}-\frac{1}{2\sqrt{2}}$

$a,b,c$ are in A.P.
$\therefore 2b=a+c...\left(1\right)$

$a^2,b^2,c^2$ are in G.P.
$\therefore (b^2{)}^2=a^2{c}^2...(2)$
$a+b+c=\frac{3}{4}...\left(3\right)$

From equation $\left(1\right)$ and $\left(3\right)$
$\Rightarrow 2b+b=\frac{3}{4}\Rightarrow b=\frac{1}{4}$
From equation $\left(2\right)$
$\Rightarrow a^2{c}^2={\left(\frac{1}{16}\right)}^2$
$\Rightarrow ac=\pm \frac{1}{16}$
$\Rightarrow a(\frac{1}{2}-a)=\pm \frac{1}{16}(\because \text{From equation}(1),a+c=\frac{1}{2})$

Case $1:$
$a(\frac{1}{2}-a)=+\frac{1}{16}$
$\Rightarrow 2a^2-a+\frac{1}{8}=0$
$\Rightarrow a=\frac{1\pm \sqrt{1-1}}{4}=\frac{1}{4}$

Case $2:$
$a(\frac{1}{2}-a)=-\frac{1}{16}$
$\Rightarrow 2a^2-a-\frac{1}{8}=0$
$\Rightarrow a=\frac{1\pm \sqrt{1+1}}{4}=\frac{1\pm \sqrt{2}}{4}$
$\Rightarrow a=\frac{1}{4}+\frac{1}{2\sqrt{2}},\frac{1}{4}-\frac{1}{2\sqrt{2}}$
$\therefore a=\frac{1}{4}-\frac{1}{2\sqrt{2}}$
$(\because a<b<c)$