Let a, b, c and d are in a geometric progression such that `a lt b lt c lt d, a + d=112` amd `b+c=48`. If the geometric progression is continued with a as the first term, then the sum of the first six terms is

Options

A. 1156
B. 1256
C. 1356
D. 1456

Answer

To solve the problem step by step, we will start by defining the terms of the geometric progression (GP) and then use the given conditions to find the values of \( a \) and \( r \).

### Step 1: Define the terms of the GP
Let the first term be \( a \) and the common ratio be \( r \). Then the terms of the GP can be expressed as:
- First term: \( a \)
- Second term: \( b = ar \)
- Third term: \( c = ar^2 \)
- Fourth term: \( d = ar^3 \)

### Step 2: Set up the equations based on the problem statement
From the problem, we have two equations:
1. \( a + d = 112 \)
2. \( b + c = 48 \)

Substituting the expressions for \( d \), \( b \), and \( c \):
1. \( a + ar^3 = 112 \) (Equation 1)
2. \( ar + ar^2 = 48 \) (Equation 2)

### Step 3: Simplify the equations
From Equation 1:
\[ a(1 + r^3) = 112 \]
From Equation 2:
\[ ar(1 + r) = 48 \]

### Step 4: Solve for \( a \) in terms of \( r \)
From Equation 2, we can express \( a \):
\[ a = \frac{48}{r(1 + r)} \]

### Step 5: Substitute \( a \) into Equation 1
Substituting \( a \) into Equation 1:
\[ \frac{48}{r(1 + r)}(1 + r^3) = 112 \]
Multiplying both sides by \( r(1 + r) \):
\[ 48(1 + r^3) = 112r(1 + r) \]

### Step 6: Expand and rearrange the equation
Expanding both sides:
\[ 48 + 48r^3 = 112r + 112r^2 \]
Rearranging gives:
\[ 48r^3 - 112r^2 - 112r + 48 = 0 \]

### Step 7: Simplify the equation
Dividing the entire equation by 16:
\[ 3r^3 - 7r^2 - 7r + 3 = 0 \]

### Step 8: Factor or use the quadratic formula
To solve the cubic equation, we can use the Rational Root Theorem or synthetic division. Testing possible rational roots, we find:
\[ (r - 3)(3r^2 + 2r - 1) = 0 \]
Thus, one root is \( r = 3 \).

### Step 9: Solve for \( a \) using \( r = 3 \)
Using \( r = 3 \) in the equation for \( a \):
\[ a = \frac{48}{3(1 + 3)} = \frac{48}{12} = 4 \]

### Step 10: Find the terms of the GP
Now we can find the terms:
- \( a = 4 \)
- \( b = ar = 4 \cdot 3 = 12 \)
- \( c = ar^2 = 4 \cdot 3^2 = 36 \)
- \( d = ar^3 = 4 \cdot 3^3 = 108 \)

### Step 11: Calculate the sum of the first six terms of the GP
The sum of the first \( n \) terms of a GP is given by:
\[ S_n = \frac{a(r^n - 1)}{r - 1} \]
For \( n = 6 \):
\[ S_6 = \frac{4(3^6 - 1)}{3 - 1} \]
Calculating \( 3^6 = 729 \):
\[ S_6 = \frac{4(729 - 1)}{2} = \frac{4 \cdot 728}{2} = 1456 \]

### Final Answer
The sum of the first six terms of the geometric progression is \( \boxed{1456} \).