Recurrences, let's start with easier

The given recurrence is $a_n=7a_{n-1}-6a_{n-2}$ . So, we first determine its characteristic polynomial in order to solve the linear recurrence. Therefore, the equation becomes

$x^2=7x-6$ , i.e.

$x^2-7x+6=0$ . Now, the roots of this polynomial are $x=6/1$

. So, according to the notation given in the reference link, we get $a_n=c_1(6)^n+c_2(1)^n$ .. Now, we find value of $c_1$ and $c_2$ by substituting $n=0,1$ . So we get the relations that

$c_1+c_2=0$ and $6c_1+c_2=1$ . Solving these equations, we get

$c_1=\frac{1}{5},c_2=\frac{-1}{5}$ Therefore, the general term is given by

$a_n=\frac{1}{5}(6^n-1)$

So, now we have to find $a_{20}$ , that is

$a_{20}=\frac{1}{5}(6^{20}-1)$ .

Therefore, our task remains just to find the last 3 digits of
$(6^{20}-1)\frac{1}{5}$ , which is easy using congruence.