Need help with this Geometric sequence problem First, sorry if Im not using the right syntax, im translating the problem and im not sure if im supposed to say "Sequence" or "series", and also thanks to who ever tries to help.
The sum of a geometric  sequence is 20, and the sum of its squared terms is 205.
find how many terms are in the sequence if the first term is $\frac{1}{2}$.
$a$1  + $a$2  + $a$3  +...+$a$$n$  = 20
$a$1 2 + $a$2 2 + $a$3 2 +...+$a$$n$ 2 = 205
$a$1 = $\frac{1}{2}$
find $n$.
$20=\frac{a\left(q^n-1\right)}{q-1}$   ==>     $205=\frac{10.25a\left(q^n-1\right)}{q-1}$
$205=\frac{a^2\left(q^{2n}-1\right)}{q^2-1}$ = $\frac{a^2\left(q^n-1\right)\left(q^n+1\right)}{\left(q-1\right)\left(q+1\right)}$
$\frac{10.25a\left(q^n-1\right)}{q-1}$ =$\frac{a^2\left(q^n-1\right)\left(q^n+1\right)}{\left(q-1\right)\left(q+1\right)}$
$10.25a\left(q^n-1\right)\left(q+1\right)$ = $a^2\left(q^n-1\right)\left(q^n+1\right)$
$20.5\left(q+1\right)\space =\space \left(q^n+1\right)$
This is basically where I tell myself i did something wrong.

You have found that $q^n+1=20.5(q+1)$. 
But since the sum of the terms is $20$, we have $q^n-1=40(q-1)$.  (In essence you had written down this equation also.)
Subtract, and find  $q$, and then $n$.  The numbers are disappointingly small, $3$ and $4$ respectively.
