Question

Suppose $a,b,c$ are in A.P. 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 of $a$ is

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

Answer 1

The correct option is D

$\frac{1}{2}-\frac{1}{\sqrt{2}}$

Explanation for correct option

Determining the value of $a$

Given, $a,b,c$ are in A.P.

$\Rightarrow 2b=a+c\dots \dots \left(1\right)$

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

$\Rightarrow b^2=\pm ac\dots \dots \left(2\right)$

and,

$a+b+c=\frac{3}{2}\dots \dots \left(3\right)$

Substituting equation $\left(1\right)$ in equation $\left(3\right)$:

we get,

$2b+b=\frac{3}{2}\dots \dots \left(4\right)$

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

Substituting $b=\frac{1}{2}$ in the above equations,

$$\begin{gathered} \left(1\right)\Rightarrow a+c=1 \\ \left(2\right)\Rightarrow ac=\pm \frac{1}{4} \end{gathered}$$

Hence, a linear equation can be formed having roots $a\&c$

${\mathrm{x}}^2-(\mathrm{sum}\mathrm{of}\mathrm{the}\mathrm{roots})\mathrm{x}+(\mathrm{product}\mathrm{of}\mathrm{the}\mathrm{roots})=0$

$x^2-x+\frac{1}{4}=0and{x}^2-x-\frac{1}{4}=0$

$4x^2-4x+1=0and4x^2-4x-1=0$

Therefore the two equations are,

$4x^2-4x+1=0\&4x^2-4x-1=0$

Solving the equations simultaneously we get,

$x=\frac{1}{2}-\frac{1}{\sqrt{2}}$ or $x=\frac{1}{2}+\frac{1}{\sqrt{2}}$

$\therefore a=\frac{1}{2}-\frac{1}{\sqrt{2}}$

Hence, the correct answer is Option (D).