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

Answer 1

Answer: (B)

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

Let a=m-d, b=m and c=m+d where d is the common difference.

Now according to the question $a,b,c$ are in $AP$

$\Rightarrow a+c=2b$

Given that,

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

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

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

Also,

$\begin{array}{cc}& a^2,b^2,c^2\text{are in}G\cdot P\text{.}\\ & \Rightarrow a^2{c}^2={\left(b^2\right)}^2\\ & \Rightarrow (m-d{)}^2(m+d{)}^2=m^4\\ & \Rightarrow {(m^2-d^2)}^2=m^4\\ & \Rightarrow m^4+d^4-2m^2{d}^2=m^2\\ & \Rightarrow d^2(d^2-2m^2)=0\\ \Rightarrow & d^2=2m^2\text{or}d=\frac{1}{\sqrt{2}}\end{array}$

$\begin{array}{cc}\text{Therefore}& a=m-d\\ \Rightarrow & a=\frac{1}{2}-\frac{1}{\sqrt{2}}\end{array}$