The product of three consecutive terms of a GP is 512. If 4 is added to each of the first and the second of these terms, the three terms now form an AP. Then the sum of the original three terms of the given GP is: (a) 36 (b) 32 (c) 24 (d) 28

Options

A. 36
B. 24
C. 32
D. 28

Answer

To solve the problem step by step, we will follow the reasoning laid out in the video transcript.

### Step 1: Define the terms of the GP
Let the three consecutive terms of the geometric progression (GP) be:
- First term: \( \frac{a}{r} \)
- Second term: \( a \)
- Third term: \( ar \)

### Step 2: Set up the equation for the product of the terms
According to the problem, the product of these three terms is given as:
\[ \frac{a}{r} \cdot a \cdot ar = 512 \]
Simplifying this, we get:
\[ \frac{a^3}{r} = 512 \]
Multiplying both sides by \( r \):
\[ a^3 = 512r \]

### Step 3: Solve for \( a \)
Since \( 512 = 8^3 \), we can express \( a^3 \) as:
\[ a^3 = 8^3 \implies a = 8 \]

### Step 4: Substitute \( a \) back into the product equation
Now substituting \( a = 8 \) back into the equation \( a^3 = 512r \):
\[ 8^3 = 512r \implies 512 = 512r \implies r = 1 \]

### Step 5: Find the terms of the GP
Now, substituting \( a = 8 \) and \( r = 1 \):
- First term: \( \frac{8}{1} = 8 \)
- Second term: \( 8 \)
- Third term: \( 8 \cdot 1 = 8 \)

Thus, the three terms of the GP are \( 8, 8, 8 \).

### Step 6: Check the condition for AP
Now, according to the problem, if we add 4 to the first and second terms, we get:
- First term: \( 8 + 4 = 12 \)
- Second term: \( 8 + 4 = 12 \)
- Third term: \( 8 \)

Now we check if \( 12, 12, 8 \) forms an arithmetic progression (AP):
The common difference between the terms should be the same:
\[ 12 - 12 = 0 \quad \text{and} \quad 12 - 8 = 4 \]
Since the common difference is not the same, we realize that we need to explore other values of \( r \).

### Step 7: Explore other possible values of \( r \)
From the product equation \( a^3 = 512r \), we can also express it as:
\[ r = \frac{a^3}{512} \]
Now we need to find the values of \( r \) such that the terms \( \frac{8}{r}, 8, 8r \) form an AP when we add 4 to the first two terms.

### Step 8: Set up the condition for AP
The terms in AP condition can be expressed as:
\[ \frac{8}{r} + 4, \quad 8 + 4, \quad 8r \]
This means:
\[ \left(8 + 4 - \frac{8}{r} - 4\right) = \left(8r - (8 + 4)\right) \]
This leads to the quadratic equation:
\[ 2r^2 - 5r + 2 = 0 \]

### Step 9: Solve the quadratic equation
Using the quadratic formula:
\[ r = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{5 \pm \sqrt{(-5)^2 - 4 \cdot 2 \cdot 2}}{2 \cdot 2} = \frac{5 \pm \sqrt{25 - 16}}{4} = \frac{5 \pm 3}{4} \]
This gives us:
\[ r = 2 \quad \text{or} \quad r = \frac{1}{2} \]

### Step 10: Calculate the terms for both values of \( r \)
1. If \( r = 2 \):
- First term: \( \frac{8}{2} = 4 \)
- Second term: \( 8 \)
- Third term: \( 8 \cdot 2 = 16 \)

2. If \( r = \frac{1}{2} \):
- First term: \( 8 \cdot 2 = 16 \)
- Second term: \( 8 \)
- Third term: \( \frac{8}{\frac{1}{2}} = 4 \)

### Step 11: Find the sum of the original three terms
In both cases, the terms are \( 4, 8, 16 \) or \( 16, 8, 4 \). The sum is:
\[ 4 + 8 + 16 = 28 \]

### Final Answer
Thus, the sum of the original three terms of the given GP is:
\[ \boxed{28} \]