Primeiro calculamos a razão:
[tex]r=a_2-a_1=\dfrac{1}{2}-2=-\dfrac{3}{2}[/tex]
Agora calculamos [tex]a_{20}[/tex]:
[tex]a_n=a_1+(n-1)r\\\\
a_{20}=2+(20-1)(-\dfrac{3}{2})\\\\
a_{20}=2+19(-\dfrac{3}{2})\\\\
a_{20}=2-\dfrac{57}{2}\\\\
a_{20}=\dfrac{4}{2}-\dfrac{57}{2}\\\\
a_{20}=-\dfrac{53}{2}\\\\[/tex]
Agora podemos calcular a soma:
[tex]S_n=\dfrac{(a_1+a_n)n}{2}\\\\
S_{20}=\dfrac{(a_1+a_{20})20}{2}\\\\
S_{20}=(2-\dfrac{53}{2})10\\\\
S_{20}=20-265\\\\
\boxed{S_{20}=-245}[/tex]