Step-by-step Solution:

Step 1: Define the GP Let the terms of the GP in ascending order be: \[ a, \, b, \, c, \, d \] Since they are in GP, the common ratio \( r \) satisfies: \[ b = a r, \quad c = a r^2, \quad d = a r^3 \] Step 2: Use the Given Conditions We are given: 1. \( a + d = 112 \), 2. \( b + c = 48 \). Substitute the GP terms into these equations: \[ a + a r^3 = 112 \quad \text{(1)} \] \[ a r + a r^2 = 48 \quad \text{(2)} \] Step 3: Simplify the Equations From equation (1): \[ a (1 + r^3) = 112 \quad \text{(1)} \] From equation (2): \[ a r (1 + r) = 48 \quad \text{(2)} \] Divide equation (1) by equation (2) to eliminate \( a \): \[ \frac{1 + r^3}{r (1 + r)} = \frac{112}{48} = \frac{7}{3} \] Simplify the left-hand side: \[ \frac{1 + r^3}{r (1 + r)} = \frac{(1 + r)(1 - r + r^2)}{r (1 + r)} = \frac{1 - r + r^2}{r} \] Thus: \[ \frac{1 - r + r^2}{r} = \frac{7}{3} \] Multiply through by \( r \): \[ 1 - r + r^2 = \frac{7}{3} r \] Multiply through by 3 to eliminate the fraction: \[ 3 - 3r + 3r^2 = 7r \] Rearrange: \[ 3r^2 - 10r + 3 = 0 \] Step 4: Solve for \( r \) Solve the quadratic equation \( 3r^2 - 10r + 3 = 0 \): \[ r = \frac{10 \pm \sqrt{(-10)^2 - 4 \cdot 3 \cdot 3}}{2 \cdot 3} = \frac{10 \pm \sqrt{100 - 36}}{6} = \frac{10 \pm \sqrt{64}}{6} = \frac{10 \pm 8}{6} \] Thus: \[ r = \frac{18}{6} = 3 \quad \text{or} \quad r = \frac{2}{6} = \frac{1}{3} \] Since the terms are in ascending order, \( r > 1 \), so \( r = 3 \). Step 5: Find \( a \) Substitute \( r = 3 \) into equation (2): \[ a \cdot 3 (1 + 3) = 48 \] \[ a \cdot 3 \cdot 4 = 48 \] \[ 12a = 48 \Rightarrow a = 4 \] Step 6: Find the Sum of the First Six Terms The sum of the first \( n \) terms of a GP is: \[ S_n = a \frac{r^n - 1}{r - 1} \] For \( n = 6 \), \( a = 4 \), and \( r = 3 \): \[ S_6 = 4 \frac{3^6 - 1}{3 - 1} = 4 \frac{729 - 1}{2} = 4 \cdot \frac{728}{2} = 4 \cdot 364 = 1456 \] Step 7: Conclusion The sum of the first six terms is: \[ \boxed{1456} \] Correct Answer: \(\boxed{D}\)