Step 1 Given: \(-9 + (-12) + (-15)+....\) It is an arithmetic series with \(a = \text{first term} = -9\)

\(f = \text{common difference} = -12 - (-9) = -3\) Now sum of the first n terms is, \(S_n = \frac{n}{2} [2a + (n - 1)d]\)

\(\Rightarrow S_n = n/2 [2(-9) + (n - 1)(-3)]\)

\(\Rightarrow S_n = \frac{n}{2} [-18 - 3 n + 3]\)

\(\Rightarrow S_n = \frac{n}{2} [-15 - 3n]\)

\(\Rightarrow S_n = \frac{3n}{2} [5+n]\)

Step 2 Given \(a = 2\ and\ a_{15} = 86\) Consider, \(a_{15} = 86\)

\(\Rightarrow a+(15-1)d = 86 [ \because a_n=a+(n-1)d]\)

\(\Rightarrow 2+14d = 86\)

\(\Rightarrow 14d = 84\)

\(\Rightarrow d = 6\) Therefore the sum of the first 100 terms is, \(S_{100} = \frac{100}{2} [2(2) + (100 - 1)(6)] [ \because S_n = \frac{n}{2} [2a + (n - 1)d]]\)

\(= 50(4 + 594)\)

\(=29900\)