The first term of a G.P. whose second term is 2 and sum to infinity is 8 will be

Options

A. 6
B. 3
C. 4
D. 1

Answer

To solve the problem, we need to find the first term \( a \) of a geometric progression (G.P.) given that the second term is 2 and the sum to infinity is 8.

### Step-by-Step Solution:

1. **Identify the terms of the G.P.**:
- Let the first term be \( a \).
- Let the common ratio be \( r \).
- The second term of the G.P. is given by \( ar \).

From the problem, we know:
\[ ar = 2 \quad \text{(1)} \]

2. **Use the formula for the sum to infinity of a G.P.**:
The sum to infinity \( S_\infty \) of a G.P. is given by the formula:
\[ S_\infty = \frac{a}{1 - r} \]
We are given that \( S_\infty = 8 \):
\[ \frac{a}{1 - r} = 8 \quad \text{(2)} \]

3. **Express \( a \) in terms of \( r \)**:
From equation (2), we can express \( a \):
\[ a = 8(1 - r) \quad \text{(3)} \]

4. **Substitute equation (3) into equation (1)**:
Substitute \( a \) from equation (3) into equation (1):
\[ 8(1 - r)r = 2 \]

5. **Simplify the equation**:
Expanding the equation gives:
\[ 8r - 8r^2 = 2 \]
Rearranging it leads to:
\[ 8r^2 - 8r + 2 = 0 \]

6. **Divide the entire equation by 2**:
This simplifies our quadratic equation:
\[ 4r^2 - 4r + 1 = 0 \]

7. **Factor the quadratic equation**:
This can be factored as:
\[ (2r - 1)^2 = 0 \]
Therefore, we find:
\[ 2r - 1 = 0 \implies r = \frac{1}{2} \]

8. **Substitute \( r \) back to find \( a \)**:
Now, substitute \( r = \frac{1}{2} \) back into equation (3) to find \( a \):
\[ a = 8(1 - \frac{1}{2}) = 8 \times \frac{1}{2} = 4 \]

### Final Answer:
The first term \( a \) of the G.P. is \( 4 \).

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