Chapter 13 - Sequences and Series - 13.1 Sequences - Exercises and Problems for Section 13.1 - Exercises and Problems - Page 539: 27

See below.

We know that $a_n=a_1+(n-1)d.$ where $d$=common difference and $a_1$= the first term Hence, here we have $a_n=6+(n-1)\cdot1.2=6+1.2n-1.2=1.2n+4.8$ Therefore $a_5=1.2\cdot5+4.8=6+4.8=10.8$ $a_{50}=1.2\cdot50+4.8=60+4.8=64.8$