Question

Let $a, b, c$ be in an AP and $a^2,b^2,c^2$ be in GP, if $a < b < c$ and $a+b+c= \frac{3}{2}$, then the value of a is

• A. $\frac{1}{2 \sqrt2}$
• B. $\frac{1}{2 \sqrt3}$
• C. $\frac{1}{2}-\frac{1}{\sqrt3}$
• D. $\frac{1}{2}-\frac{1}{\sqrt2}$

Answer 1

The Correct Option is D

Since, a,b and c are in an AP.
Let $\hspace40mm a =A - D, b= A, c = A+D$
Given, $\hspace20mm a + b + c = \frac{3}{2} $
$\Rightarrow \, \, (A -D) + A + (A + D) =\frac{3}{2}$
$\Rightarrow \hspace35mm 3A =\frac{3}{2} \, \, \Rightarrow \, \, \, A =\frac{1}{2}$
$\therefore$ The number are$ \frac{1}{2} -D,\frac{1}{2},\frac{1}{2}+D.$
Also, $\bigg(\frac{1}{2}-D\bigg)^2, \frac{1}{4}, \bigg(\frac{1}{2}-D\bigg)^2$ are in Gp.
$\therefore$ $ \hspace25mm \bigg(\frac{1}{4}\bigg)^2 = \bigg(\frac{1}{2}-D\bigg)^2\, \bigg(\frac{1}{2}+D\bigg)^2 $
$\Rightarrow \hspace30mm \frac{1}{16} = \bigg(\frac{1}{4}-D^2\bigg)^2\, $
$\Rightarrow \hspace22mm \frac{1}{4} -D^2 = \pm \frac{1}{4} $
$\Rightarrow \hspace30mm D^2 = \frac{1}{2} \Rightarrow D = \pm \frac{1}{\sqrt 2}$
$\therefore$ $\hspace30mm a = \frac{1}{2} \pm \frac{1}{\sqrt 2}$
So, out of the given values, $ a = \frac{1}{2} - \frac{1}{\sqrt 2}$ is the right choice.