If a,b,c are in A.P. and a2,b2,c2 are in G.P. such that a

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

1. A

$\frac{1}{4}-\frac{1}{\sqrt{2}}$

2. B

$\frac{1}{4}-\frac{1}{3\sqrt{2}}$

3. C

$\frac{1}{4}-\frac{1}{4\sqrt{2}}$

4. D

$\frac{1}{4}-\frac{1}{2\sqrt{2}}$

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### Solution:

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

Again, ${a}^{2},{b}^{2},{c}^{2}$ are in G.P.

…(ii)

But ${b}^{2}=ac$ is not possible.

($\therefore$This gives $a=b=c=1/4\right)$

Also, $a+b+c=3/4⇒3b=3/4⇒b=1/4$

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

…(iv)

From (iii), we have 1/2 $-a=c$

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

$\begin{array}{l}⇒a=\frac{8±\sqrt{64+64}}{2×16}=\frac{8±8\sqrt{2}}{2×16}⇒a=\frac{1±\sqrt{2}}{4}=\frac{1}{4}±\frac{1}{2\sqrt{2}}\\ ⇒a=\frac{1}{4}-\frac{1}{2\sqrt{2}}\end{array}$  Register to Get Free Mock Test and Study Material

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