Find the 20th term in the following sequence whose nth term is $a_{n}=\frac{n(n-2)}{n+3} ; a_{20}$
Substituting $n=20$, we obtain
$a_{20}=\frac{20(20-2)}{20+3}=\frac{20(18)}{23}=\frac{360}{23}$
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