Let A be the first sequence, and denote by [tex]a_n[/tex] the n-th term in A.
Consider the forward differences of A :
16 - 7 = 9
30 - 16 = 14
49 - 30 = 19
73 - 49 = 24
Call this sequence of first differences B, so that for n ≥ 1,
[tex]b_n=a_{n+1}-a_n[/tex]
where [tex]b_n[/tex] is the n-th term of B.
Now consider the forward differences of B, which is another sequence we'll call C :
14 - 9 = 5
19 - 14 = 5
24 - 19 = 5
Then if [tex]c_n[/tex] is the n-th term of C, we have for all n ≥ 1,
[tex]c_n=b_{n+1}-b_n=5[/tex]
which gives
[tex]b_{n+1} = b_n+5[/tex]
This tells us B is an arithmetic sequence - the first term is 9 and the difference between consecutive terms is 5, so we have for n ≥ 1,
[tex]b_n = 9 + 5(n-1) = 5n + 4[/tex]
Plug this into the recurrence for A :
[tex]a_{n+1} = a_n + b_n = a_n + 5n + 4[/tex]
We don't need to solve for [tex]a_n[/tex], fortunately; we just want the next term, which would be
[tex]a_6 = a_5 + 5^2 + 4 = 73 + 25 + 4 = \boxed{102}[/tex]
