Let a (n) denote the n-th term of the sequence. Since the terms form an arithmetic sequence, you have
a (n) = a (n - 1) + d
for some fixed constant d. You can write any term in the sequence as a function of the first term:
a (n) = [a (n - 2) + d] + d = a (n - 2) + 2d
a (n) = [a (n - 3) + d] + 2d = a (n - 3) + 3d
and so on, down to
a (n) = a (1) + (n - 1) d
so given that a (11) = 57 (so that n = 11), you have
a (1) + 10d = 57 … … … [1]
The sum of the first and fourth terms is 29:
a (1) + a (4) = 29
but we can write a (4) in terms of a (1) :
a (1) + [a (1) + 3d] = 29
which gives us another equation in a (1) and d :
2 a (1) + 3d = 29 … … … [2]
Now solve [1] and [2] :
-2 (a (1) + 10d) + (2 a (1) + 3d) = -2 (57) + 29
-17d = -85
d = 5
a (1) + 10d = a (1) + 50 = 57
a (1) = 7
Then the n-th term of the sequence is given by
a (n) = 7 + 5 (n - 1) = 5n + 2
