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