$\begin{array}{1 1}\pm 2 \\ \pm 3 \\ \pm 9 \\ \pm 6 \end{array} $

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- $n^{th}$ term of a G.P. $=t_n=a.r^{n-1}$

Given: $1^{st}$ term =1 and sum of $3^{rd}$ and $5^{th}$$= 90$

$\Rightarrow\:a=1\:\: and \:\: t_3+t_5=90$

We know that $n^{th}$ term of a G.P. is $t_n=a.r^{n-1}$

$\Rightarrow\:t_3=a.r^{2-1}=1\times r^2=r^2$ and

$t_5=a.r^{5-1}=1\times r^4=r^4$.

$\therefore\:r^2+r^4=90$

Let $r^2=x$

then $x+x^2=90$

$\Rightarrow\:x^2+x-90=0$

$\Rightarrow\:x^2+10x-9x-90=0$

$\Rightarrow\:x(x+10)-9(x+10)=0$

$\Rightarrow\:(x-9)(x+10)=0$

$\Rightarrow\:x=-10\:\:or\:\:x=9$

But since $r^2$ cannot be negative, $x=r^2=9$

$\Rightarrow\:r=\pm3$

$i.e.,$ The common ratio $=\pm 3$

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