$\frac{1}{50}$
$\frac{1}{49}$
$\frac{1}{48}$
$\frac{1}{51}$

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$5a_na_{n+1}=a_n-a_{n+1}$

$\Rightarrow 5a_na_{n+1}+a_{n+1} = a_n$

$\Rightarrow a_{n+1} = \frac{a_n}{5a_n+1}$

$\Rightarrow a_{n+1} = \frac{1}{5+\frac{1}{a_n}}$

Substituting a few values,

$a_2 = \frac{1}{6}, a_3 = \frac{1}{11}, a_4 = \frac{1}{16}$

We see that the denominator is an Arithmetic Progression.

Hence,

The series would go on like this : $1,\frac{1}{6},\frac{1}{11},\frac{1}{16}\ldots$

We find $a_{11}$ by simple observation as,

$a_{11} = \frac{1}{51}$