<s>[INST] 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 $a<b<c$   ?
 [/INST] $2b=a+c,\;
a+b+c=\dfrac{3}{2}$
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
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