Question

Suppose $a,b,c$ 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 the value is

• A. $\frac{1}{2\sqrt{2}}$
• B. $\frac{1}{2\sqrt{3}}$
• C. $\frac{1}{2}-\frac{1}{\sqrt{3}}$
• D. $\frac{1}{2}-\frac{1}{\sqrt{2}}$

Answer 1

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

Since, $a,b,c$ are in $AP$.
$\therefore a=A-D,b=A,c=A+D$
Where, $A$ is the first term and $D$ is the common difference of an $AP$.
Given, $a+b+c=\frac{3}{2}$
$\Rightarrow (A-D)+A+(A+D)=\frac{3}{2}$
$\Rightarrow 3A=\frac{3}{2}$
$\Rightarrow A=\frac{1}{2}$
$\therefore$ The number are $\frac{1}{2}-D,\frac{1}{2},\frac{1}{2}+D$
Also, ${(\frac{1}{2}-D)}^2,\frac{1}{4},{(\frac{1}{2}+D)}^2$ are in GP.
$\Rightarrow {\left(\frac{1}{4}\right)}^2={(\frac{1}{2}-D)}^2{(\frac{1}{2}+D)}^2$
$\Rightarrow \frac{1}{16}={(\frac{1}{4}-D^2)}^2$
$\Rightarrow \frac{1}{4}-D^2=\pm \frac{1}{4}$
$\Rightarrow D^2=\frac{1}{2}(D=0$ is not possible)
$\Rightarrow D=\pm \frac{1}{\sqrt{2}}$
Therefore, $a=\frac{1}{2}\pm \frac{1}{\sqrt{2}}$
So, out of the given value $a=\frac{1}{2}-\frac{1}{\sqrt{2}}$ is the right choice.