If a, b, c are in A.P. and a^2, b^2, c^2 are | vedclass

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(D) Since $a, b, c$ are in $A.P.$,we have $a+c = 2b$.Given $a+b+c = \frac{3}{4}$,substituting $a+c = 2b$ gives $3b = \frac{3}{4}$,so $b = \frac{1}{4}$

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

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(D) Since $a, b, c$ are in $A.P.$,we have $a+c = 2b$.Given $a+b+c = \frac{3}{4}$,substituting $a+c = 2b$ gives $3b = \frac{3}{4}$,so $b = \frac{1}{4}$.Since $a^2, b^2, c^2$ are in $G.P.$,we have $(b^2)^2 = a^2c^2$,which implies $ac = \pm b^2 = \pm \frac{1}{16}$.Since $a We have $a+c = 2b = \frac{1}{2}$ and $ac = -\frac{1}{16}$.These are roots of the quadratic equation $x^2 - (a+c)x + ac = 0$,which is $x^2 - \frac{1}{2}x - \frac{1}{16} = 0$.Multiplying by $16$,we get $16x^2 - 8x - 1 = 0$.Using the quadratic formula $x = \frac{-B \pm \sqrt{B^2 - 4AC}}{2A}$,we get $x = \frac{8 \pm \sqrt{64 - 4(16)(-1)}}{32} = \frac{8 \pm \sqrt{128}}{32} = \frac{8 \pm 8\sqrt{2}}{32} = \frac{1}{4} \pm \frac{\sqrt{2}}{4} = \frac{1}{4} \pm \frac{1}{2\sqrt{2}}$.Since $a < b$,we choose the smaller root: $a = \frac{1}{4} - \frac{1}{2\sqrt{2}}$.

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

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