# 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 $a<b<c$ ?

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

$2b=a+c,\; a+b+c=\dfrac{3}{2}$

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

$b=qa , c=q^2 a$ so we have $$2q=1+q^2$$ hence $q=1.$ Therefore $a=b=c=0.5$ is a unique solution.

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