<s>[INST] What is the value of $c$ in this problem of sequence and series? $a,b,c$  is in A.P. and $a^2,b^2,c^2$ is in G.P.If $a<b<c$ & $a + b + c = \dfrac {3}{2}$ ,what is the value of $c$?[When I wanted to do it,I found $a + d = \dfrac {1}{2}$,$d$ being the common difference,say.Thus $b$ = $\dfrac{1}{2}$  . Now I get $a + c = 1$.As $a^2,b^2,c^2$ is in G.P. we get $b$ =$\sqrt{ac}$ 
& finally get $a = c$ But it will be a contradiction.]
 [/INST] Note that the geometric progression is in the squares of $a,b,c$ so that you have $a^2c^2=b^4$ whence $ac=\pm b^2$
You have $a+c=1$ from the AP condition and $ac=\pm b^2=\pm\frac 14$ from the GP condition. With the positive sign we know that $a,c$ are roots of the quadratic $$x^2-x+\frac 14=\left(x-\frac 12\right)^2=0$$Which is the case you have considered. But with the negative sign:$$x^2-x-\frac 14=0=\left(x-\frac 12\right)^2-\frac 12$$ and $x=\cfrac {1\pm \sqrt 2}{2}$ giving the arithmetic progression $$\frac {1- \sqrt 2}2,\frac 12, \frac {1+ \sqrt 2}2$$ with common difference $\cfrac {\sqrt 2}2$ and the geometric progression $$\frac {3-2\sqrt 2}4, \frac 14, \frac {3+2\sqrt 2}4$$ with common ratio $3+2\sqrt 2$
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