The sum of an infinite numbers of a G.P. is 2 and the sum of their cubes is 24. The second term of the G.P. is

Options

A. `-(1)/(2)`
B. `(1)/(2)`
C. 2
D. `-(3)/(2)`

Answer

To solve the problem, we need to find the second term of a geometric progression (G.P.) given that the sum of the infinite terms is 2 and the sum of their cubes is 24.

### Step-by-Step Solution:

1. **Understanding the Sum of an Infinite G.P.**:
The sum \( S \) of an infinite G.P. is given by the formula:
\[ S = \frac{A}{1 - r} \]
where \( A \) is the first term and \( r \) is the common ratio. According to the problem, we have:
\[ \frac{A}{1 - r} = 2 \quad \text{(1)} \]

2. **Understanding the Sum of Cubes of the G.P.**:
The sum of the cubes of the terms in the G.P. can be expressed as:
\[ S_c = \frac{A^3}{1 - r^3} \]
Given that this sum is 24, we have:
\[ \frac{A^3}{1 - r^3} = 24 \quad \text{(2)} \]

3. **Expressing \( 1 - r^3 \)**:
We can express \( 1 - r^3 \) in terms of \( 1 - r \):
\[ 1 - r^3 = (1 - r)(1 + r + r^2) \]

4. **Substituting into Equation (2)**:
Substituting \( 1 - r^3 \) into equation (2):
\[ \frac{A^3}{(1 - r)(1 + r + r^2)} = 24 \]
From equation (1), we know \( 1 - r = \frac{A}{2} \). Substituting this into the equation gives:
\[ \frac{A^3}{\frac{A}{2}(1 + r + r^2)} = 24 \]
Simplifying this, we get:
\[ \frac{2A^2}{1 + r + r^2} = 24 \]
Therefore:
\[ 2A^2 = 24(1 + r + r^2) \quad \text{(3)} \]

5. **Using Equation (1) to Find \( A \)**:
From equation (1), we can express \( A \):
\[ A = 2(1 - r) \]
Substituting \( A \) into equation (3):
\[ 2(2(1 - r))^2 = 24(1 + r + r^2) \]
Simplifying gives:
\[ 8(1 - r)^2 = 24(1 + r + r^2) \]
Dividing both sides by 8:
\[ (1 - r)^2 = 3(1 + r + r^2) \]

6. **Expanding and Rearranging**:
Expanding both sides:
\[ 1 - 2r + r^2 = 3 + 3r + 3r^2 \]
Rearranging gives:
\[ 0 = 2r^2 + 5r + 2 \quad \text{(4)} \]

7. **Factoring the Quadratic Equation**:
Factoring equation (4):
\[ (2r + 1)(r + 2) = 0 \]
This gives us:
\[ r = -\frac{1}{2} \quad \text{or} \quad r = -2 \]

8. **Finding Corresponding Values of \( A \)**:
For \( r = -\frac{1}{2} \):
\[ A = 2(1 - (-\frac{1}{2})) = 2 \times \frac{3}{2} = 3 \]
For \( r = -2 \):
\[ A = 2(1 - (-2)) = 2 \times 3 = 6 \]

9. **Finding the Second Term**:
The second term of the G.P. is given by \( Ar \):
- For \( r = -\frac{1}{2} \):
\[ \text{Second term} = 3 \times -\frac{1}{2} = -\frac{3}{2} \]
- For \( r = -2 \):
\[ \text{Second term} = 6 \times -2 = -12 \]

10. **Conclusion**:
The second term of the G.P. can be either \(-\frac{3}{2}\) or \(-12\). However, since the problem asks for the second term of the G.P. and the options are typically limited, we take \(-\frac{3}{2}\) as the correct answer.

### Final Answer:
The second term of the G.P. is \(-\frac{3}{2}\).