If $\alpha ,\beta$  be the roots of $x^2-x-1=0$  and $A_n={\alpha }^n+{\beta }^n$ , then AM of $A_{n-1}$  and $A_n$  is

The given equation is $x^2-x-1=0$

Let $\alpha ,\beta$  are the roots given equation

$\therefore$  Sum of the roots : $\alpha +\beta =1$  and

     Product of the roots : $\alpha \beta =-1$

Given that $A_n={\alpha }^n+{\beta }^n$

AM of $A_{n-1}$  and $A_n=\frac{A_{n-1}+A_n}{2}$

                $=\frac{{\alpha }^{n-1}+{\beta }^{n-1}+{\alpha }^n+{\beta }^n}{2}$

                $=\frac{{\alpha }^{n-1}\left(1+\alpha \right)+{\beta }^{n-1}\left(1+\beta \right)}{2}$

               $=\frac{{\alpha }^{n-1}\left({\alpha }^2\right)+{\beta }^{n-1}\left({\beta }^2\right)}{2}$          $\left(\because {\alpha }^2=\alpha +1\text{and}{\beta }^2=\beta +1\right)$

               $=\frac{1}{2}\left({\alpha }^{n+1}+{\beta }^{n+1}\right)$

               $=\frac{1}{2}{A}_{n+1}$