Consecutive and twice!

First of all I would use sine rule which states that $\dfrac{sin\angle A}{a}=\dfrac{sin\angle B}{b}=\dfrac{sin\angle C}{c}$

now let sides be $x-1,x,x+1$ where longest side is $x+1$ and shortest is $x-1$

$\Rightarrow \dfrac{sin\angle 2a}{x+1}=\dfrac{sin\angle a}{x-1}=\dfrac{sin\angle (180-3a)}{x}$

$\Rightarrow \dfrac{sin\angle 2a}{x+1}=\dfrac{sin\angle a}{x-1} \Rightarrow (x-1)sin 2a=(x+1)(sin a)$

$\Rightarrow 2(x-1)sin(a)cos(a)=(x+1)sin(a)$

$\Rightarrow cos(a)=\dfrac{x+1}{2x-2}$

now apply cosine rule

$\implies$ $(x-1)^2=x^2+(x+1)^2-2(x)(x+1)(cos(a)) \\ \Rightarrow x^2+1-2x=x^2+x^2+1+2x-2(x^2+x)(\dfrac{x+1}{2x-2}) \\ \Rightarrow x^2+1-2x=x^2+x^2+1+2x-2x^2-2x)(\dfrac{x+1}{2x-2})$

$\Rightarrow 4x+x^2=\dfrac{-2[(x+1)^2(x)}{2x-2} \\ \Rightarrow \dfrac{2(x+1)^2}{2x-2}=4+x \\ \Rightarrow 2x^2+2+4x=8x-8+2x^2 -2x \\ \Rightarrow 2x=10 \therefore x=5$

now according to question $x-1=n \therefore n=4$

So our answer is $\huge\Rightarrow \color{royalblue}{\boxed 4}$

3,4,5 could be the sides coz angle 45° and 90°