## Algebra 2 (1st Edition)

n-th term:$\quad a_{n}=3.1+(n-1)0.7$, The next term is $\quad a_{5}=5.9$
Note that there is a common difference of 0.7 between neighboring terms. $n=1,\quad 3.1=3.1+0$ $n=2,\quad 3.8=3.1+0.7\quad =3.1+(2-1)0.7$ $n=3,\quad 4.5=3.8+0.7=3.1+2\times 0.7\quad =3.1+(3-1)0.7$ $n=4,\quad 5.2=4.5+0.7=3.1+3\times 0.7\quad =3.1+(4-1)0.7$ So, the nth term is $a_{n}=3.1+(n-1)0.7$, and the next term is $a_{5}=5.2+0.7=5.9$ (also, $3.1+(4)0.7=5.9$)