The largest term in the sequence $a_n=\frac{n}{n^2+100}$ is:

$a_{10}$

$a_n=\frac{1}{n+100/n}$

$a_n$ is maximum if $n+\frac{100}{n}$ is minimum.

Now, $E=n+\frac{100}{n}⇒\frac{dE}{dn}=1-\frac{100}{n^2}$

$\frac{dE}{dn}=0⇒\frac{100}{n^2}=1⇒n^2=100⇒n=±10$

Now, $\frac{d^2E}{dn^2}=1+\frac{200}{n^3}⇒(\frac{d^2E}{dn^2})_{n=10}=1+\frac{200}{1000}>0$ ⇒ E has minimum when n = 10

⇒ $a_n$ is maximum when n = 10 ∴ The largest term of sequence is $a_{10}$.