1. **State the problem:** We are given the 2nd and 5th terms of a geometric progression (GP) as $-6$ and $48$ respectively. We need to find the sum of the first four terms. 2. **Recall the formula for the $n$th term of a GP:** $$a_n = a r^{n-1}$$ where $a$ is the first term and $r$ is the common ratio. 3. **Write equations for the given terms:** - 2nd term: $a_2 = a r = -6$ - 5th term: $a_5 = a r^4 = 48$ 4. **Divide the 5th term by the 2nd term to find $r^3$:** $$\frac{a r^4}{a r} = \frac{48}{-6} \Rightarrow r^3 = -8$$ 5. **Solve for $r$:** $$r = \sqrt[3]{-8} = -2$$ 6. **Find $a$ using the 2nd term:** $$a r = -6 \Rightarrow a \times (-2) = -6 \Rightarrow a = \frac{-6}{-2} = 3$$ 7. **Find the first four terms:** - $a_1 = a = 3$ - $a_2 = a r = 3 \times (-2) = -6$ - $a_3 = a r^2 = 3 \times (-2)^2 = 3 \times 4 = 12$ - $a_4 = a r^3 = 3 \times (-2)^3 = 3 \times (-8) = -24$ 8. **Sum of the first four terms:** $$S_4 = a + a r + a r^2 + a r^3 = 3 - 6 + 12 - 24 = -15$$ **Final answer:** The sum of the first four terms is $-15$.