We are required to find the first 4 terms of the sequence an = 63 +(n-1)5
The nth term of the sequence is given by the formula
an = 63 +(n-1)5
When n = 1
[tex]\begin{gathered} a_1=63+(1-1)5\text{ = 63 + (0)5= 63+0} \\ a_1=63 \end{gathered}[/tex]The first term is 63
when n=2
[tex]\begin{gathered} a_2=63+(2-1)5\text{ = 63 + (1)5= 63+}5 \\ a_2=68 \end{gathered}[/tex]The second term is 68
When n=3
[tex]\begin{gathered} a_3=63+(3-1)5\text{ = 63 + (2)5= 63+}10 \\ a_3=73 \end{gathered}[/tex]The third term is 73
When n = 4
[tex]\begin{gathered} a_4=63+(4-1)5\text{ = 63 + (3)5= 63+}15 \\ a_4=78 \end{gathered}[/tex]The fouth term is 78
Hence, the first 4 terms of the sequence are 63, 68, 73, 78
