Let a(n) denote the n-th term in the sequence. Because the terms are in arithmetic progression, there is a fixed number d that separates consecutive terms, so that starting with a(1) = a, the next few terms are
a(2) = a(1) + d = a + d
a(3) = a(2) + d = a + 2d
a(4) = a(3) + d = a + 3d
and so on, up to
a(n) = a + (n - 1) d
We're given that the 30th term is 98, so
a(30) = a + 29d = 98
The sum of the first 30 terms is 1635, so that
[tex]\displaystyle \sum_{n=1}^{30}a(n) = \sum_{n=1}^{30}(a+(n-1)d) \\\\ 1635 = a\sum_{n=1}^{30}1 + d\sum_{n=1}^{30}(n-1) \\\\ 1635 = 30a + d\sum_{n=0}^{29}n \\\\ 1635 = 30a + d\sum_{n=1}^{29}n \\\\ 1635 = 30a + \frac{d\times29\times30}2[/tex]
so that
30a + 435d = 1635
Solve the equations in boldface for a and d. I'll eliminate a and solve for d first.
-30 (a + 29d) + (30a + 435d) = -30 (98) + 1635
-30a - 870d + 30a + 435d = -2940 + 1635
-435d = -1305
d = 3
Then
a + 29 (3) = 98
a + 87 = 98
a = 11
