The sum of an infinite geometric progression (G.P.) is 2 and the sum of the G.P. made from the cubes of the terms of this infinite series is 24. The values a nad r respectively (where is the first term and r denote common ratio of the series)

Options

A. A. `-2,3`
B. B. `3, -1/2`
C. C. `0, 1/2`
D. D. can't be determined

Answer

To solve the problem, we need to find the values of \( a \) (the first term) and \( r \) (the common ratio) of the infinite geometric progression (G.P.) given the following conditions:

1. The sum of the infinite G.P. is 2.
2. The sum of the G.P. formed by the cubes of the terms of this series is 24.

### Step-by-Step Solution:

**Step 1: Use the formula for the sum of an infinite G.P.**

The formula for the sum \( S \) of an infinite G.P. is given by:

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

Given that the sum is 2, we have:

\[ \frac{a}{1 - r} = 2 \quad \text{(1)} \]

**Step 2: Use the formula for the sum of the cubes of the terms of the G.P.**

The sum of the cubes of the terms of the G.P. can be expressed as:

\[ S_{\text{cubes}} = \frac{a^3}{1 - r^3} \]

Given that this sum is 24, we have:

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

**Step 3: Express \( a \) in terms of \( r \) from equation (1).**

From equation (1), we can express \( a \):

\[ a = 2(1 - r) \quad \text{(3)} \]

**Step 4: Substitute \( a \) into equation (2).**

Substituting equation (3) into equation (2):

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

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

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

**Step 5: Simplify the equation.**

Cross-multiplying gives:

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

Dividing both sides by 8:

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

**Step 6: Expand both sides.**

Expanding \( (1 - r)^3 \):

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

Expanding the right-hand side:

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

**Step 7: Rearranging the equation.**

Bringing all terms to one side:

\[ 0 = 3r^3 - r^3 + 3r^2 - 3r + 1 - 3 \]

This simplifies to:

\[ 2r^3 + 3r^2 - 3r - 2 = 0 \quad \text{(4)} \]

**Step 8: Solve the cubic equation (4).**

Using the Rational Root Theorem or synthetic division, we can find the roots of the cubic equation. Testing \( r = -1/2 \):

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

Calculating each term:

\[ 2(-1/8) + 3(1/4) + 3/2 - 2 = -1/4 + 3/4 + 3/2 - 2 = 0 \]

Thus, \( r = -1/2 \) is a root.

**Step 9: Find \( a \) using the value of \( r \).**

Substituting \( r = -1/2 \) back into equation (3):

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

### Final Answer:

The values of \( a \) and \( r \) are:

\[ a = 3, \quad r = -\frac{1}{2} \]