The sum of the sequence of three distinct real numbers, which are in GP is  $S^2$. If their sum is $\alpha S$ , then ${\alpha }^2\in$.

Let three terms be $a,ar,a{r}^2$

Given $a+ar+a{r}^2=\alpha S$                                         ... (i)

and $a^2+a^2{r}^2+a^2{r}^4=S^2$                                     ..(ii)

Dividing (i) and (ii), we get

$\begin{array}{l}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\frac{{\left(1+r+r^2\right)}^2}{\left(1+r^2+r^4\right)}=\frac{{\alpha }^2{S}^2}{S^2}={\alpha }^2\\ \Rightarrow \hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\frac{\left(1+r+r^2\right)}{\left(1-r+r^2\right)}={\alpha }^2\\ \Rightarrow \hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}{\alpha }^2\left(1-r+r^2\right)=\left(1+r+r^2\right)\\ \Rightarrow \hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\left({\alpha }^2-1\right)r^2-\left({\alpha }^2+1\right)r+\left({\alpha }^2-1\right)=0\\ \text{ As }r\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\text{ is real, so }{\left({\alpha }^2+1\right)}^2-4{\left({\alpha }^2-1\right)}^2\ge 0\\ \Rightarrow \hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}{\left({\alpha }^2+1\right)}^2-{\left(2{\alpha }^2-2\right)}^2\ge 0\\ \Rightarrow \hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\left({\alpha }^2+1+2{\alpha }^2-2\right)\left({\alpha }^2+1-2{\alpha }^2+2\right)\ge 0\\ \Rightarrow \hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\left(3{\alpha }^2-1\right)\left(-{\alpha }^2+3\right)\ge 0\\ \Rightarrow \hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\left(3{\alpha }^2-1\right)\left({\alpha }^2-3\right)\le 0\\ \Rightarrow \hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\frac{1}{3}\le {\alpha }^2\le 3\end{array}$

But  ${\alpha }^2=1\Rightarrow r=0$

It is not possible. 

Thus, $\frac{1}{3}\le {\alpha }^2\le 3-\left\{1\right\}$

$\Rightarrow {\alpha }^2\in \left(\frac{1}{3},1\right)\cup (1,3)$

Hence, the result.