Browse Questions

# Find $a_{20}$ if $a_n=\large\frac{n(n-2)}{n+3}$

$\begin{array}{1 1}\frac{360}{23} \\ \frac{240}{23} \\ \frac{380}{23} \\ \frac{420}{23} \end{array}$

$a_n=\large\frac{n(n-2)}{n+3}$
By putting $n=20$ we get the $20^{th}$ term as
$a_{20}=\large\frac{20(20-2)}{20+3}=\frac{360}{23}$