Chapter 12 - Sequences; Induction; the Binomial Theorem - 12.2 Arithmetic Sequences - 12.2 Assess Your Understanding - Page 814: 30

$140\sqrt5.$

We know that $a_n=a_1+(n-1)d.$ where $d$=common difference and $a_1$= the first term Here $a_1=2\sqrt5$ and $d=a_2-a_1=4\sqrt5-2\sqrt5=2\sqrt5$ Hence, here we have $a_n=2\sqrt5+(n-1)\cdot2\sqrt5.$ Therefore $a_{70}=2\sqrt5+(70-1)\cdot2\sqrt5\\=140\sqrt5.$