Chapter 13 - Sequences and Series - 13.1 Sequences - Exercises and Problems for Section 13.1 - Exercises and Problems - Page 539: 25

$101$

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.