The largest term of the sequence $<a_n>$ given by $a_n=\frac{n^2}{n^3+200}, n \in N$, is

$\frac{49}{543}$

Let $f(n)=a_n=\frac{n^2}{n^3+200}, n \in N$. Then,

$f'(n)=\frac{2\left(n^3+200\right) n-3 n^4}{n^3+200}=\frac{n\left(400-n^3\right)}{\left(n^3+200\right)^2}$

Clearly, $f'(n) \neq 0$ for any $n \in N$. So, $f(n)$ has no extreme point.

But,

$f'(n)>0$ for $n \leq 7$ and, $f'(n)<0$ for $n \geq 8$

$\Rightarrow f(n)$ is increasing for $n \leq 7$ and decreasing for $n \geq 8$.

Now, $f(7)=\frac{49}{543}$ and $f(8)=\frac{8}{89}$

Clearly, $f(7)>f(8)$.

Hence, the value of the largest term is $\frac{49}{543}$.