The sum of three numbers in G.P. is 14 . If the first two terms are each increased by 1 and the third term is decreased by 1, the resulting numbers are in A.P. Find the product of these three numbers :

Options

A. 125
B. 64
C. 216
D. 124

Answer

To solve the problem step by step, let's break it down:

### Step 1: Define the terms in G.P.
Let the three numbers in Geometric Progression (G.P.) be:
- First term = \( a \)
- Second term = \( ar \)
- Third term = \( ar^2 \)

### Step 2: Write the equation for their sum
According to the problem, the sum of these three numbers is 14:
\[ a + ar + ar^2 = 14 \]
Factoring out \( a \):
\[ a(1 + r + r^2) = 14 \quad \text{(Equation 1)} \]

### Step 3: Modify the terms to form an A.P.
When we increase the first two terms by 1 and decrease the third term by 1, we get:
- New first term = \( a + 1 \)
- New second term = \( ar + 1 \)
- New third term = \( ar^2 - 1 \)

These new terms are in Arithmetic Progression (A.P.), which means:
\[ 2(ar + 1) = (a + 1) + (ar^2 - 1) \]
Simplifying this equation:
\[ 2ar + 2 = a + 1 + ar^2 - 1 \]
\[ 2ar + 2 = a + ar^2 \]
Rearranging gives us:
\[ ar^2 - 2ar + a - 2 = 0 \quad \text{(Equation 2)} \]

### Step 4: Solve the equations
Now we have two equations:
1. \( a(1 + r + r^2) = 14 \)
2. \( ar^2 - 2ar + a - 2 = 0 \)

From Equation 1, we can express \( a \):
\[ a = \frac{14}{1 + r + r^2} \]

### Step 5: Substitute \( a \) into Equation 2
Substituting \( a \) into Equation 2:
\[ \frac{14r^2}{1 + r + r^2} - 2\frac{14r}{1 + r + r^2} + \frac{14}{1 + r + r^2} - 2 = 0 \]
Multiplying through by \( 1 + r + r^2 \) to eliminate the denominator:
\[ 14r^2 - 28r + 14 - 2(1 + r + r^2) = 0 \]
Expanding gives:
\[ 14r^2 - 28r + 14 - 2 - 2r - 2r^2 = 0 \]
Combining like terms:
\[ 12r^2 - 30r + 12 = 0 \]
Dividing through by 6:
\[ 2r^2 - 5r + 2 = 0 \]

### Step 6: Factor or use the quadratic formula
Using the quadratic formula \( r = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
\[ r = \frac{5 \pm \sqrt{(-5)^2 - 4 \cdot 2 \cdot 2}}{2 \cdot 2} \]
\[ r = \frac{5 \pm \sqrt{25 - 16}}{4} \]
\[ r = \frac{5 \pm 3}{4} \]
This gives us two possible values for \( r \):
1. \( r = 2 \)
2. \( r = \frac{1}{2} \)

### Step 7: Find corresponding values of \( a \)
1. For \( r = 2 \):
\[ a(1 + 2 + 4) = 14 \implies a \cdot 7 = 14 \implies a = 2 \]
The numbers are \( 2, 4, 8 \).

2. For \( r = \frac{1}{2} \):
\[ a(1 + \frac{1}{2} + \frac{1}{4}) = 14 \implies a \cdot \frac{7}{4} = 14 \implies a = 8 \]
The numbers are \( 8, 4, 2 \).

### Step 8: Calculate the product
In both cases, the product of the numbers is:
\[ 2 \times 4 \times 8 = 64 \]

### Final Answer
The product of the three numbers is \( \boxed{64} \).