Question

If $a, b, c$ are in $A.P$. and $a^2, b^2, c^2$ are in $G.P$. such that $a < b < c $ and $a + b + c = \frac{3}{4}$, then the value of $a$ is :

• A. $\frac{1}{4} - \frac{1}{ 4 \sqrt{2}}$
• B. $\frac{1}{4} - \frac{1}{3 \sqrt{2}}$
• C. $\frac{1}{4} - \frac{1}{2 \sqrt{2}}$
• D. $\frac{1}{4} - \frac{1}{\sqrt{2}}$

Answer 1

Answer: (C)

Given $a, b, c$ in $AP$
$\Rightarrow 2 b=a+c\,\,\,\,\,\,\,(1)$
And $a^{2}, b^{2}, c^{2}$ in GP
$\Rightarrow b^{4}=a^{2} c^{2} $
$\Rightarrow b^{2}=\pm a c\,\,\,\,\,\,\,(2)$
Taking $b^{2}=-a c\,\,\,\,\,\,\,$(From Eq. (1))
$\left(\frac{a+c}{2}\right)^{2} =-a c$
$a^{2}+c^{2}+2 \,a c =-4 \,a c $
$ a^{2}+c^{2}+6\, a c =0 \,\,\,\,\,\,\,(3)$
$a+b+c =\frac{3}{4}$(Given)
$\Rightarrow \frac{a+c}{2}+(a+c) =\frac{3}{4} \,\,\,\,\,\,\,$(From Eq. (1))
$ \Rightarrow a+c =\frac{1}{2} \,\,\,\,\,\,\,(4)$
$\Rightarrow a^{2}+c^{2} =\frac{1}{4}-2 \,a c \,\,\,\,\,\,\,$(Squaring above Eq. (4))
$ \frac{1}{4}-2 \,a c+6\, a c =0$ (From Eq. (3))
$ \Rightarrow a c =\frac{-1}{16}\,\,\,\,\,\,\,(5)$
$\Rightarrow b^{2} =\frac{1}{16} $
$\Rightarrow b=\frac{1}{4}, \frac{-1}{4} $
$\Rightarrow a\left(\frac{1}{2}-a\right) =\frac{-1}{16} $
$\Rightarrow a^{2}-\frac{a}{2}-\frac{1}{16}=0 \,\,\,\,\,\,\,$(From Eq. (1) and Eq. (5))
$\Rightarrow a=\frac{\frac{1}{2} \pm \sqrt{\frac{1}{4}+\frac{1}{4}}}{2}=\frac{1}{4} \pm \frac{1}{2 \sqrt{2}}$
$b > a \Rightarrow a=\frac{1}{4}-\frac{1}{2 \sqrt{2}}$