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