Chapter 12 Sequences and Series - 12.2 Analyze Arithmetic Sequences and Series - 12.2 Exercises - Skill Practice - Page 806: 12

$a_n=-2+3n$ ; $a_{20}= 58$

We know that $a_n= a_1+(n-1) d$ Here, we have $d=4-1=3$ $a_n= 1+(n-1) \times 3$ This gives: $a_n=-2+3n$ when $n=20$, we have $a_{20}= -2+(20) \times 3=58$ Hence, $a_n=-2+3n$ ; $a_{20}= 58$