The sum of an infinite geometric series is 2 and the sum of the geometric series made from the cubes of this infinite series is 24. Then the series is:

Options

A. `3,2/3,-3/4,3/8`…..
B. `3,3/2,2/4,3/8`,…..
C. `3,-3/2,3/4,-3/8`,….
D. `1,-1/2,1/4,-1/8`,…..

Answer

To solve the problem step by step, we will follow the reasoning presented in the video transcript while ensuring clarity and completeness.

### Step 1: Set Up the Infinite Geometric Series
Let the first term of the infinite geometric series be \( a \) and the common ratio be \( r \). The sum of an infinite geometric series is given by the formula:

\[ S = \frac{a}{1 - r} \]

According to the problem, the sum of the series is 2:

\[ \frac{a}{1 - r} = 2 \]

### Step 2: Express \( a \) in Terms of \( r \)
From the equation above, we can express \( a \):

\[ a = 2(1 - r) \]

Let’s label this as Equation (1).

### Step 3: Set Up the Series of Cubes
The series made from the cubes of the terms of the original series is:

\[ a^3 + (ar)^3 + (ar^2)^3 + \ldots \]

This can be expressed as:

\[ \frac{a^3}{1 - r^3} = 24 \]

### Step 4: Substitute \( a \) into the Cubic Series Equation
Substituting \( a = 2(1 - r) \) into the cubic series sum gives:

\[ \frac{(2(1 - r))^3}{1 - r^3} = 24 \]

Calculating \( (2(1 - r))^3 \):

\[ = \frac{8(1 - r)^3}{1 - r^3} = 24 \]

### Step 5: Simplify the Equation
Now, we can simplify this equation:

\[ 8(1 - r)^3 = 24(1 - r^3) \]

Dividing both sides by 8:

\[ (1 - r)^3 = 3(1 - r^3) \]

### Step 6: Expand and Rearrange
Expanding \( (1 - r)^3 \):

\[ 1 - 3r + 3r^2 - r^3 = 3(1 - r^3) \]

Expanding the right side:

\[ 1 - 3r + 3r^2 - r^3 = 3 - 3r^3 \]

Rearranging gives:

\[ 2r^3 + 3r^2 - 3r - 2 = 0 \]

### Step 7: Factor the Polynomial
We can check for rational roots. Testing \( r = 1 \):

\[ 2(1)^3 + 3(1)^2 - 3(1) - 2 = 0 \]

Thus, \( r - 1 \) is a factor. We can factor the polynomial as:

\[ (r - 1)(2r^2 + 5r + 2) = 0 \]

### Step 8: Solve the Quadratic Equation
Now we solve \( 2r^2 + 5r + 2 = 0 \) using the quadratic formula:

\[ r = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{-5 \pm \sqrt{25 - 16}}{4} = \frac{-5 \pm 3}{4} \]

Calculating the roots:

1. \( r = \frac{-2}{4} = -\frac{1}{2} \)
2. \( r = \frac{-8}{4} = -2 \)

### Step 9: Determine Valid \( r \)
Since \( |r| < 1 \), the valid solution for \( r \) is \( r = -\frac{1}{2} \).

### Step 10: Find \( a \)
Substituting \( r = -\frac{1}{2} \) back into Equation (1):

\[ a = 2(1 - (-\frac{1}{2})) = 2(1 + \frac{1}{2}) = 2 \cdot \frac{3}{2} = 3 \]

### Step 11: Write the Series
Now we can write the series:

\[ 3, \quad 3(-\frac{1}{2}), \quad 3(-\frac{1}{2})^2, \quad 3(-\frac{1}{2})^3, \ldots \]

This gives us:

\[ 3, \quad -\frac{3}{2}, \quad \frac{3}{4}, \quad -\frac{3}{8}, \ldots \]

### Final Answer
The series is:

\[ 3, -\frac{3}{2}, \frac{3}{4}, -\frac{3}{8}, \ldots \]