If the second term of GP is 2 and the sum of its infinite terms is 8, then the GP is:

Options

A. A)`8,2,(1)/(2),(1)/(8)` . . .
B. B)`10,2,(2)/(5),(2)/(25)`
C. C)`4,2,1,(1)/(2),(1)/(2^(2))` . . .
D. D)`6,3,(3)/(2),(3)/(4)`

Answer

To solve the problem, we need to find the geometric progression (GP) given the second term and the sum of its infinite terms. Let's break it down step by step.

### Step 1: Define the terms of the GP
Let the first term of the GP be denoted as \( A \) and the common ratio as \( r \).

### Step 2: Use the information about the second term
The second term of the GP can be expressed as:
\[ \text{Second term} = A \cdot r \]
According to the problem, the second term is equal to 2:
\[ A \cdot r = 2 \quad \text{(1)} \]

### Step 3: Use the information about the sum of infinite terms
The sum of the infinite terms of a GP is given by the formula:
\[ S = \frac{A}{1 - r} \]
According to the problem, the sum of the infinite terms is equal to 8:
\[ \frac{A}{1 - r} = 8 \quad \text{(2)} \]

### Step 4: Solve the equations simultaneously
From equation (1), we can express \( A \) in terms of \( r \):
\[ A = \frac{2}{r} \quad \text{(3)} \]
Now, substitute equation (3) into equation (2):
\[ \frac{\frac{2}{r}}{1 - r} = 8 \]
This simplifies to:
\[ \frac{2}{r(1 - r)} = 8 \]
Cross-multiplying gives:
\[ 2 = 8r(1 - r) \]
Expanding this, we have:
\[ 2 = 8r - 8r^2 \]
Rearranging gives us a quadratic equation:
\[ 8r^2 - 8r + 2 = 0 \]
Dividing the entire equation by 2 simplifies it to:
\[ 4r^2 - 4r + 1 = 0 \]

### Step 5: Solve the quadratic equation
Using the quadratic formula \( r = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
Here, \( a = 4, b = -4, c = 1 \):
\[ r = \frac{4 \pm \sqrt{(-4)^2 - 4 \cdot 4 \cdot 1}}{2 \cdot 4} \]
Calculating the discriminant:
\[ r = \frac{4 \pm \sqrt{16 - 16}}{8} = \frac{4 \pm 0}{8} = \frac{4}{8} = \frac{1}{2} \]

### Step 6: Find the value of \( A \)
Now substituting \( r = \frac{1}{2} \) back into equation (3):
\[ A = \frac{2}{\frac{1}{2}} = 4 \]

### Step 7: Write the GP
Now we have \( A = 4 \) and \( r = \frac{1}{2} \). The GP can be written as:
\[ 4, \quad 4 \cdot \frac{1}{2} = 2, \quad 4 \cdot \left(\frac{1}{2}\right)^2 = 1, \quad 4 \cdot \left(\frac{1}{2}\right)^3 = \frac{1}{2}, \quad \ldots \]
Thus, the GP is:
\[ 4, 2, 1, \frac{1}{2}, \frac{1}{4}, \ldots \]

### Final Answer
The GP is \( 4, 2, 1, \frac{1}{2}, \frac{1}{4}, \ldots \)

---