Question

Suppose $a,b,c$ are in $AP$ and $a^2,b^2,c^2$ are in GP. If $a<b<c$ and $a+b+c=\frac{3}{2}$, then the value of $a$ 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

Answer: (D)

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 numbers are $\frac{1}{2}-D,\frac{1}{2},\frac{1}{2}+D$
Also, ${\left(\frac{1}{2}-D\right)}^2,\frac{1}{4},{\left(\frac{1}{2}+D\right)}^2$ are in GP.
$\Rightarrow {\left(\frac{1}{4}\right)}^2={\left(\frac{1}{2}-D\right)}^2{\left(\frac{1}{2}+D\right)}^2$
$\Rightarrow \frac{1}{16}={\left(\frac{1}{4}-D^2\right)}^2$
$\Rightarrow \frac{1}{4}-D^2=\pm \frac{1}{4}$
$\Rightarrow D^2=(\frac{1}{2}D=0$ 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.