If I'm reading this correctly, the table is telling you that [tex]a_4=16[/tex] and [tex]a_{37}=115[/tex].
Because the sequence follows an arithmetic progression, there is a common difference [tex]d[/tex] between terms such that
[tex]a_n=a_{n-1}+d[/tex]
You can solve for [tex]a_n[/tex] explicitly in terms of the first term [tex]a_1[/tex] by recursively substituting the right side:
[tex]a_n=a_{n-1}+d[/tex]
[tex]\implies a_n=a_{n-2}+2d[/tex]
[tex]\implies a_n=a_{n-3}+3d[/tex]
[tex]\implies\cdots\implies a_n=a_1+(n-1)d[/tex]
Let's first find the common difference. You have
[tex]a_5=a_4+d[/tex]
[tex]\implies a_6=a_5+d=a_4+2d[/tex]
[tex]\implies a_7=a_6+d=a_5+2d=a_4+3d[/tex]
[tex]\implies\cdots\implies a_{37}=a_{36}+d=\cdots=a_4+33d[/tex]
Substituting in the terms of the sequence that you know, you get
[tex]115=16+33d\implies d=3[/tex]
Now you can find the first term:
[tex]a_4=a_1+(4-1)d\implies 16=a_1+9\implies a_1=7[/tex]
and the 22nd term:
[tex]a_{22}=a_1+(22-1)d=70[/tex]
