<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] $2b=a+c$ 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:
$$(a,b,c)=\left(\frac{1-\sqrt2}{2},\frac{1}{2},\frac{1+\sqrt2}{2}\right)$$
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