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

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(D) Let the terms in $A.P.$ be $a = b - d$ and $c = b + d$,where $d > 0$ since $a < b < c$.Given $a + b + c = \frac{3}{2}$,we have $(b - d) + b + (b +

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

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(D) Let the terms in $A.P.$ be $a = b - d$ and $c = b + d$,where $d > 0$ since $a Given $a + b + c = \frac{3}{2}$,we have $(b - d) + b + (b + d) = \frac{3}{2}$,which implies $3b = \frac{3}{2}$,so $b = \frac{1}{2}$.Thus,$a = \frac{1}{2} - d$ and $c = \frac{1}{2} + d$.Since $a^2, b^2, c^2$ are in $G.P.$,we have $(b^2)^2 = a^2 c^2$,which means $b^4 = (ac)^2$.Taking the square root,$b^2 = \pm ac$.If $b^2 = ac$,then $b^2 = (b - d)(b + d) = b^2 - d^2$,so $d^2 = 0$,which implies $d = 0$. This contradicts $a Therefore,$b^2 = -ac$.Substituting $b = \frac{1}{2}$,$a = \frac{1}{2} - d$,and $c = \frac{1}{2} + d$:$(\frac{1}{2})^2 = -(\frac{1}{2} - d)(\frac{1}{2} + d)$$\frac{1}{4} = -(\frac{1}{4} - d^2)$$\frac{1}{4} = -\frac{1}{4} + d^2$$d^2 = \frac{1}{2}$,so $d = \frac{1}{\sqrt{2}}$ (since $d > 0$).Finally,$a = \frac{1}{2} - d = \frac{1}{2} - \frac{1}{\sqrt{2}}$.

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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