1. **State the problem:** We have an arithmetic series where the sum of the first 48 terms is 4 times the sum of the first 36 terms. We need to find the sum of the first 30 terms. 2. **Recall the formula for the sum of the first n terms of an arithmetic series:** $$S_n = \frac{n}{2} [2a + (n-1)d]$$ where $a$ is the first term and $d$ is the common difference. 3. **Set up the equation given:** $$S_{48} = 4 S_{36}$$ Using the formula: $$\frac{48}{2} [2a + 47d] = 4 \times \frac{36}{2} [2a + 35d]$$ 4. **Simplify the equation:** $$24 (2a + 47d) = 72 (2a + 35d)$$ 5. **Expand both sides:** $$48a + 1128d = 144a + 2520d$$ 6. **Bring all terms to one side:** $$48a - 144a + 1128d - 2520d = 0$$ $$-96a - 1392d = 0$$ 7. **Divide entire equation by -96:** $$\cancel{-96}a + \cancel{-96} \times 14.5 d = 0 \Rightarrow a + 14.5 d = 0$$ 8. **Express $a$ in terms of $d$:** $$a = -14.5 d$$ 9. **Find the sum of the first 30 terms:** $$S_{30} = \frac{30}{2} [2a + 29d] = 15 (2a + 29d)$$ 10. **Substitute $a = -14.5 d$ into the sum:** $$S_{30} = 15 [2(-14.5 d) + 29 d] = 15 (-29 d + 29 d) = 15 \times 0 = 0$$ **Final answer:** $$\boxed{0}$$