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

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(D) Given $a, b, c$ are in $A.P.$,let $b = a + d$ and $c = a + 2d$,where $d > 0$ since $a < b < c$.Given $a^2, b^2, c^2$ are in $G.P.$,we have $(b^2)^

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

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(D) Given $a, b, c$ are in $A.P.$,let $b = a + d$ and $c = a + 2d$,where $d > 0$ since $a Given $a^2, b^2, c^2$ are in $G.P.$,we have $(b^2)^2 = a^2 c^2$,which implies $b^4 = a^2 c^2$.Taking the square root,$b^2 = \pm ac$.If $b^2 = ac$,then $a, b, c$ are in $G.P.$ Since they are also in $A.P.$,$a = b = c$,which contradicts $a Thus,$b^2 = -ac$.Substituting $b = a + d$ and $c = a + 2d$:$(a + d)^2 = -a(a + 2d)$$a^2 + d^2 + 2ad = -a^2 - 2ad$$2a^2 + 4ad + d^2 = 0$.Given $a + b + c = \frac{3}{2}$,we have $a + (a + d) + (a + 2d) = 3a + 3d = \frac{3}{2}$,so $a + d = \frac{1}{2}$,which means $d = \frac{1}{2} - a$.Substituting $d$ into the equation $2a^2 + 4ad + d^2 = 0$:$2a^2 + 4a(\frac{1}{2} - a) + (\frac{1}{2} - a)^2 = 0$$2a^2 + 2a - 4a^2 + \frac{1}{4} - a + a^2 = 0$$-a^2 + a + \frac{1}{4} = 0$$4a^2 - 4a - 1 = 0$.Using the quadratic formula $a = \frac{-(-4) \pm \sqrt{(-4)^2 - 4(4)(-1)}}{2(4)} = \frac{4 \pm \sqrt{16 + 16}}{8} = \frac{4 \pm 4\sqrt{2}}{8} = \frac{1}{2} \pm \frac{1}{\sqrt{2}}$.Since $d = \frac{1}{2} - a > 0$,we must have $a Therefore,$a = \frac{1}{2} - \frac{1}{\sqrt{2}}$.

Suppose $a, b, c$ are in $A.P.$ and $a^2, b^2, c^2$ are in $G.P.$ If $a < b < c$ and $a + b + c = \frac{3}{2}$,then the value of $a$ is

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