Mathematics - Sequence & Series Question with Solution | TestHub

$\because a,b,c$ are in A.P. then $$a+c=2b$$ also it is given that, $$\begin{array}{rl}& a+b+c=\frac{3}{4}\\ & \Rightarrow 2b+b=\frac{3}{4}\Rightarrow b=\frac{1}{4}\end{array}$$ Again it is given that, $a^2,b^2,c^2$ are in G.P. then $${\left(b^2\right)}^2=a^2{c}^2\Rightarrow ac=\pm \frac{1}{16}$$ From (1), (2) and (3), we get; $$a\pm \frac{1}{16a}=\frac{1}{2}\Rightarrow 16a^2-8a\pm 1=0$$ Case I: $16a^2-8a+1=0$ $\Rightarrow a=\frac{1}{4}$ (not possible as $a<b)$ Case II: $16a^2-8a-1=0\Rightarrow a=\frac{8\pm \sqrt{128}}{32}$ $$\begin{array}{rl}& \Rightarrow a=\frac{1}{4}\pm \frac{1}{2\sqrt{2}}\\ & \therefore a=\frac{1}{4}-\frac{1}{2\sqrt{2}}(\because a<b)\end{array}$$