To solve the problem, we need to find the first six terms of a geometric progression (GP) given that \( a, b, c, d \) are in GP, \( a + d = 112 \), and \( b + c = 48 \).
### Step-by-Step Solution:
1. **Define the terms of the GP**:
Since \( a, b, c, d \) are in geometric progression, we can express them in terms of \( a \) and the common ratio \( r \):
- \( b = ar \)
- \( c = ar^2 \)
- \( d = ar^3 \)
2. **Set up the equations**:
From the problem, we have two equations:
- \( a + d = 112 \) can be rewritten as:
\[
a + ar^3 = 112 \quad \text{(1)}
\]
- \( b + c = 48 \) can be rewritten as:
\[
ar + ar^2 = 48 \quad \text{(2)}
\]
3. **Factor the equations**:
From equation (1):
\[
a(1 + r^3) = 112
\]
From equation (2):
\[
ar(1 + r) = 48
\]
4. **Express \( a \) in terms of \( r \)**:
From equation (2), we can express \( a \):
\[
a = \frac{48}{r(1 + r)} \quad \text{(3)}
\]
5. **Substitute \( a \) from equation (3) into equation (1)**:
Substitute \( a \) from equation (3) into equation (1):
\[
\frac{48}{r(1 + r)}(1 + r^3) = 112
\]
6. **Clear the fraction**:
Multiply both sides by \( r(1 + r) \):
\[
48(1 + r^3) = 112r(1 + r)
\]
7. **Expand and simplify**:
Expanding both sides:
\[
48 + 48r^3 = 112r + 112r^2
\]
Rearranging gives:
\[
48r^3 - 112r^2 - 112r + 48 = 0
\]
8. **Divide the equation by 16**:
Simplifying the equation:
\[
3r^3 - 7r^2 - 7r + 3 = 0
\]
9. **Factor the polynomial**:
Using the Rational Root Theorem or synthetic division, we can find the roots:
The polynomial can be factored as:
\[
(3r - 1)(r - 3) = 0
\]
This gives us:
\[
r = \frac{1}{3} \quad \text{or} \quad r = 3
\]
10. **Select the appropriate value for \( r \)**:
Since \( a < b < c < d \), we choose \( r = 3 \).
11. **Find \( a \)**:
Substitute \( r = 3 \) back into equation (3):
\[
a = \frac{48}{3(1 + 3)} = \frac{48}{12} = 4
\]
12. **Find the terms of the GP**:
Now we can find the terms:
- \( a = 4 \)
- \( b = ar = 4 \times 3 = 12 \)
- \( c = ar^2 = 4 \times 3^2 = 36 \)
- \( d = ar^3 = 4 \times 3^3 = 108 \)
13. **Calculate the sum of the first six terms**:
The first six terms of the GP are:
\[
4, 12, 36, 108, 324, 972
\]
The sum \( S_n \) of the first \( n \) terms of a GP is given by:
\[
S_n = a \frac{r^n - 1}{r - 1}
\]
For \( n = 6 \):
\[
S_6 = 4 \frac{3^6 - 1}{3 - 1} = 4 \frac{729 - 1}{2} = 4 \frac{728}{2} = 4 \times 364 = 1456
\]
### Final Answer:
The sum of the first six terms is \( \boxed{1456} \).
