The sequence is arithmetic, so there is a constant difference d between consecutive terms such that
[tex]a_{100} = a_{99} + d[/tex]
[tex]a_{100} = (a_{98} + d) + d = a_{98} + 2d[/tex]
[tex]a_{100} = (a_{97} + d) + 2d = a_{97} + 3d[/tex]
and so on, down to
[tex]a_{100} = a_1 + 99d[/tex]
(notice the pattern of 100 = 99 + 1 = 98 + 2 = 97 + 3 = … = 1 + 99)
Solve for d :
[tex]307 = 15 + 99d \implies 99d = 292 \implies d = \dfrac{292}{99}[/tex]
Now, the sum of the first 100 terms of this sequence is
[tex]\displaystyle \sum_{n=1}^{100} a_n = \sum_{n=1}^{100} \left(15 + \frac{292}{99}(n-1)\right) = \boxed{16100}[/tex]
which follows from the well-known sums
[tex]\displaystyle \sum_{n=1}^N 1 = N[/tex]
[tex]\displaystyle \sum_{n=1}^N n = \frac{N(N+1)}2[/tex]
