Answer
$101$
Work Step by Step
We know that $a_n=a_1+(n-1)d.$
where $d$=common difference and $a_1$= the first term
Hence, here we have
$a_n=10+(n-1)\cdot10=10+10n-10=10n-5$
Therefore $10n-5\gt1000\\10n\gt1005\\n\gt100.5$
Thus $n=101$ is the first good number.