Question

Let $a,b,c$ are in AP and $a^2,b^2,c^2$ are in G.P. If $a<b<c$ and $a+b+c=\frac{3}{2}$ then $a=$

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

Answer 1

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

$a,b,c$ are in A.P.
$\Rightarrow 2b=a+c$
$\because a+b+c=\frac{3}{2}$
$\Rightarrow 3b=\frac{3}{2}\Rightarrow b=\frac{1}{2}$
So,terms in A.P are $a=\frac{1}{2}-d,b=\frac{1}{2},c=\frac{1}{2}+d$
$\because a^2,b^2,c^2$ are in G.P.
$\Rightarrow b^4=a^2{c}^2$
$\Rightarrow ac=\pm b^2$
$\Rightarrow ac=\pm \frac{1}{4}\Rightarrow \frac{1}{4}-d^2=\pm \frac{1}{4}$

$\Rightarrow d=0,\pm \frac{1}{\sqrt{2}}$
$\Rightarrow d=\frac{1}{\sqrt{2}}(\because a<b<c)$
$\therefore a=\frac{1}{2}-\frac{1}{\sqrt{2}}$.