Question

Suppose a, b, c are in AP and $a^2,b^2,c^2$ are in GP and $a<b<c$ 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

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

Given a,b,c are in A.P then $2b=a+c$

but $a+b+c=\frac{3}{2}$

$\Rightarrow 3b=\frac{3}{2}\Rightarrow b=\frac{1}{2}$

Let the common difference of the A.P. be $d$.

Now, $a^2,b^2,c^2$ are in G.P

$\Rightarrow {(b-d)}^2,b^2,{(b+d)}^2$ are in G.P

$\Rightarrow b^4={(b-d)}^2{(b+d)}^2$

$\Rightarrow {\left\{(b-d)(b+d)\right\}}^2={(b^2-d^2)}^2$

$\Rightarrow (b^2-d^2)=\pm b^2$

Now since $+$ve sign will give $d=0$ so we will take $-$ve sign.

hence $(b^2-d^2)=-b^2$

$\Rightarrow 2b^2=d^2$

$\Rightarrow d=\pm b\sqrt{2}$

Putting the value of $b$

$d=\pm \frac{1}{\sqrt{2}}$

Since $a,b$ and $c$ are in increasing order according to question.

So, we will take $+$ve sign in $d$

Hence $a=b-d=\frac{1}{2}-\frac{1}{\sqrt{2}}$

hence the correct option is D