The sum of an infinite geometirc progression is 2 and the sum of the geometric progression made from the cubes of this infinite series is 24. Then find its first term and common ratio :-

Answer

To solve the problem, we will follow these steps:

### Step 1: Set up the equations based on the given information.
We know that the sum \( S \) of an infinite geometric progression (GP) 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:
\[ \frac{a}{1 - r} = 2 \quad \text{(1)} \]

### Step 2: Set up the second equation for the sum of the cubes.
The sum of the cubes of the terms of the GP also forms a GP, where the first term is \( a^3 \) and the common ratio is \( r^3 \). The sum of this new GP is:
\[ \frac{a^3}{1 - r^3} = 24 \quad \text{(2)} \]

### Step 3: Solve equation (1) for \( a \).
From equation (1), we can express \( a \) in terms of \( r \):
\[ a = 2(1 - r) \quad \text{(3)} \]

### Step 4: Substitute equation (3) into equation (2).
Now, substitute \( a \) from 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 side:
\[ 1 - 3r + 3r^2 - r^3 = 3 - 3r^3 \]

### Step 7: Rearranging the equation.
Rearranging gives:
\[ 1 - 3r + 3r^2 - r^3 - 3 + 3r^3 = 0 \]
Combining like terms:
\[ 2r^3 + 3r^2 - 3r - 2 = 0 \quad \text{(4)} \]

### Step 8: Factor or find roots of the polynomial.
To solve the cubic equation (4), we can use the Rational Root Theorem or synthetic division. Testing \( r = 1 \):
\[ 2(1)^3 + 3(1)^2 - 3(1) - 2 = 0 \]
So \( r - 1 \) is a factor. Dividing \( 2r^3 + 3r^2 - 3r - 2 \) by \( r - 1 \) gives:
\[ 2r^2 + 5r + 2 = 0 \]

### Step 9: Solve the quadratic equation.
Using the quadratic formula:
\[ r = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{-5 \pm \sqrt{25 - 16}}{4} = \frac{-5 \pm 3}{4} \]
This gives:
\[ r = -\frac{1}{2} \quad \text{or} \quad r = -2 \]
Since the common ratio must be between -1 and 1 for convergence, we take:
\[ r = -\frac{1}{2} \]

### Step 10: Substitute \( r \) back to find \( a \).
Using equation (3):
\[ a = 2(1 - (-\frac{1}{2})) = 2(1 + \frac{1}{2}) = 2 \times \frac{3}{2} = 3 \]

### Final Answer:
The first term \( a \) is 3 and the common ratio \( r \) is -\(\frac{1}{2}\).