Arithmetic and Geometric progression in 3 numbers Suppose 
$$a,b,c \textrm{ is an arithmetic progression}$$ and
$$a^2,b^2,c^2\textrm{ is a geometric progression}$$
$$a+b+c = \frac{3}{2}.$$
From these equations I get
$$2b=a+c  \textrm{, from A.P.}$$
$$b^4=a^2c^2 \textrm{, from G.P.}$$
and finally  $$a=b=c=\frac{1}{2}.$$
Can there be any such triplet such that $a<b<c$   ?
 A: $2b=a+c$ and $a+b+c=\frac{3}{2}$ gives $b=\frac{1}{2}$ and $a+c=1$.
Also we have $b^4=a^2c^2$, which gives $$a^2c^2=\frac{1}{16}.$$
Thus, $$a^2(1-a)^2=\frac{1}{16},$$
which gives $a(1-a)=\frac{1}{4}$ and $a=c=\frac{1}{2}$ or
$$a(1-a)=-\frac{1}{4},$$
which gives $$(a,c)\in\left\{\left(\frac{1+\sqrt2}{2},\frac{1-\sqrt2}{2}\right),\left(\frac{1-\sqrt2}{2},\frac{1+\sqrt2}{2}\right)\right\}.$$
For $a<b<c$ we have the following unique solution:
$$(a,b,c)=\left(\frac{1-\sqrt2}{2},\frac{1}{2},\frac{1+\sqrt2}{2}\right)$$
A: $2b=a+c,\;
a+b+c=\dfrac{3}{2}$
give $3b=\dfrac{3}{2}\to b=\dfrac{1}{2}$
and $a+c=1\to c=1-a$
Substitute in $b^4=a^2c^2$ and get
$a^2(1-a)^2=\dfrac{1}{16}$
$a(1-a)=\pm \dfrac{1}{4}$
$a (1-a)=\dfrac{1}{4}$ gives $a=b=c=\dfrac{1}{2}$ you have already found
$a(1-a)=-\dfrac{1}{4}$ gives 
$\color{red}{a=\dfrac{1}{2} \left(1-\sqrt{2}\right),\;b=\dfrac{1}{2};\;c=\dfrac{1}{2} \left(1+\sqrt{2}\right)}$ which is the solution you was looking for $a<b<c$
the other one
$a=\dfrac{1}{2} \left(1+\sqrt{2}\right),b=\dfrac{1}{2},c=\dfrac{1}{2} \left(1-\sqrt{2}\right)$ does not satisfy the request
Hope this is useful
A: $b=qa , c=q^2 a $ so we have $$2q=1+q^2 $$ hence $q=1. $ Therefore $a=b=c=0.5$ is a unique solution.
A: If $a^2,b^2,c^2$are in geometric progression,
Then
 $$\Big(\frac{a}{b}\Big)^2=\Big(\frac{b}{c}\Big)^2$$
$$\color{red}{1. }$$
$$\frac{a}{b}=\frac{b}{c}$$
Hence $a,b,c$  are also in GP.
Let $a,b,c$ be $f,fg,fg^2$
For $a,b,c$ to be in AP,
$$f+fg^2=2fg$$
$$g^2+1=2g$$
$$(g-1)^2=0$$
$$g=1$$
If g=1 then $$a=b=c$$ hence there is no possibility for$$a\neq b\neq c$$
$$\color{red}{2. }$$
$$\frac{a}{b}=-\frac{b}{c}$$
$$b^2=a(-c)$$
Hence a,b,-c are in GP, 
Let them be $f,fg,fg^2$
So now a,b,c are in AP,
$$a+c=2b$$
$$f-fg^2=2fg$$
$$g^2+2g-1=0$$
$$(g+1)^2=2$$
$$g=\sqrt{2} - 1 \ , or\,  -(\sqrt{2} + 1)$$
Hence $a,b,-c$ are,
$$f,(\sqrt{2}-1)f,(3-2\sqrt{2})f$$
Or
$$f,-(\sqrt{2}+1)f,(3+2\sqrt{2})f$$
For $a,b,c$ to be in AP
$$f-3f+2\sqrt{2}f=2(\sqrt{2}-1)f$$
$$2(\sqrt{2}-1)f=2(\sqrt{2}-1)f$$
Or
$$f-3f-\sqrt{2}f=-2(\sqrt{2}+1)f$$
$$-2(\sqrt{2}f+1)f=-2(\sqrt{2}+1)f$$
Which is true, hence such pairs to exist

In you case, alternate solutions are,
  $$f+(\sqrt{2}-1)f-3f+2\sqrt{2}f=\frac{3}{2}$$
  $$f(3\sqrt{2}-3)=\frac{3}{2}$$
  $$f=\frac{\sqrt{2}+1}{2}$$
  Hence one of the possible outcomes studying your equations is,
  $$\frac{\sqrt{2}+1}{2},\frac{1}{2},\frac{1-\sqrt{2}}{2}$$

