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 prove 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

The correct option is C $\frac{1}{4}-\frac{1}{2\sqrt{2}}$

Here a,b,c are in AP

Let $a=\alpha -d$

$b=\alpha$

$c=\alpha +d$ where d is the common difference

Since $a+b+c=\frac{3}{4}$

$\Rightarrow \alpha -d+\alpha +\alpha +d=\frac{3}{4}$

$\Rightarrow 3\alpha =\frac{3}{4}$

$\Rightarrow \alpha =\frac{1}{4}$

$\therefore a=\frac{1}{4}-d,b=\frac{1}{4},\alpha =\frac{1}{4}+d$

now, $a^2,b^2,c^2$ are in GP.

Let $a^2=\frac{{\alpha }^{\prime }}{\mu },b^2={\alpha }^{\prime }$ and $c^2={\alpha }^{\prime }\mu$

since $b^2={\alpha }^{\prime }$

$\Rightarrow {\alpha }^{\prime }=(\frac{1}{4}{)}^2=\frac{1}{16}$

$\therefore a^2=\frac{1}{16\mu },b^2=\frac{1}{16},c^2=\frac{1}{16}\mu$

now, $a^2{c}^2=\frac{1}{16\mu }.\frac{1}{16}\mu$

$\Rightarrow (\frac{1}{4}-d{)}^2(\frac{1}{4}+d{)}^2=(\frac{1}{16}{)}^2$

$\Rightarrow {\left\{\right(\frac{1}{4}-d\left)\right(\frac{1}{4}+d\left)\right\}}^2=(\frac{1}{16}{)}^2$

$\Rightarrow (\frac{1}{16}-d^2{)}^2=(\frac{1}{16}{)}^2$

$\Rightarrow (\frac{1}{16}{)}^2+d^4-2.\frac{1}{6}{d}^2=(\frac{1}{16}{)}^2$

$\Rightarrow d^4-\frac{1}{8}{d}^2=0$

$\Rightarrow d^2(d^2-\frac{1}{8})=0$

$\Rightarrow d^2=0$ or $d^2=\frac{1}{8}$

$\Rightarrow d=\frac{1}{2\sqrt{2}}.$

Hence $a=\frac{1}{4}-\frac{1}{2\sqrt{2}}.$