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

Solution:

Given that, $a,b,c$ care in A.P. $\Rightarrow 2b=a+c\dots \left(\mathrm{i}\right)$

Again, $a^2,b^2,c^2$ are in G.P.

$\begin{array}{l}\Rightarrow b^4=(ac{)}^2\Rightarrow \left({\left(b^2\right)}^2-(ac{)}^2\right)=0\\ \Rightarrow \left(b^2-ac\right)\left(b^2+ac\right)=0\\ \Rightarrow b^2=ac\text{ or }{b}^2=-ac\end{array}$           …(ii)

But $b^2=ac$ is not possible.

($\therefore$This gives $a=b=c=1/4)$

Also, $a+b+c=3/4\Rightarrow 3b=3/4\Rightarrow b=1/4$

Putting this value of $b$in (i) and (ii), we have

$1/2=a+c\text{...(iii) }\text{ and }1/16=-ac$         …(iv)

From (iii), we have 1/2 $-a=c$

Putting the value of $c$ in (iv), we have $16a^2-8a-1=0$

$\begin{array}{l}\Rightarrow a=\frac{8\pm \sqrt{64+64}}{2\times 16}=\frac{8\pm 8\sqrt{2}}{2\times 16}\Rightarrow a=\frac{1\pm \sqrt{2}}{4}=\frac{1}{4}\pm \frac{1}{2\sqrt{2}}\\ \Rightarrow a=\frac{1}{4}-\frac{1}{2\sqrt{2}}\end{array}$