Let `b_(1),b_(2),.......` be a gemoertic sequence such that `b_(1)+b_(2)=1and sum_(k=1)^(oo)b_(k)=2`. Given that `b_(2)lt0` then the value of `b_(1)` is :
`(a) 2-sqrt2 " "(b) 1+sqrt2 " "(c) 2+sqrt2 " "(d) 4-sqrt2 `

Options

A. `2-sqrt2`
B. `1+sqrt2`
C. `2+sqrt2`
D. `4-sqrt2`

Explanation

To solve the problem, we need to find the value of \( b_1 \) given the conditions of the geometric sequence. Let's break it down step by step.

### Step 1: Understand the Geometric Sequence
Let the first term of the geometric sequence be \( b_1 = a \) and the common ratio be \( r \). Thus, the terms can be expressed as:
- \( b_1 = a \)
- \( b_2 = ar \)

### Step 2: Set Up the Given Equations
From the problem, we have two equations:
1. \( b_1 + b_2 = 1 \)
2. \( \sum_{k=1}^{\infty} b_k = 2 \)

Substituting the expressions for \( b_1 \) and \( b_2 \) into the first equation:
\[ a + ar = 1 \]
Factoring out \( a \):
\[ a(1 + r) = 1 \quad \text{(Equation 1)} \]

### Step 3: Use the Sum of the Infinite Series
The sum of an infinite geometric series is given by:
\[ \sum_{k=1}^{\infty} b_k = \frac{a}{1 - r} \]
Setting this equal to 2:
\[ \frac{a}{1 - r} = 2 \quad \text{(Equation 2)} \]

### Step 4: Solve for \( a \) in Terms of \( r \)
From Equation 1, we can express \( a \) in terms of \( r \):
\[ a = \frac{1}{1 + r} \]

### Step 5: Substitute \( a \) into Equation 2
Substituting \( a \) into Equation 2:
\[ \frac{\frac{1}{1 + r}}{1 - r} = 2 \]
This simplifies to:
\[ \frac{1}{(1 + r)(1 - r)} = 2 \]
Cross-multiplying gives:
\[ 1 = 2(1 + r)(1 - r) \]
Expanding the right side:
\[ 1 = 2(1 - r^2) \]
This simplifies to:
\[ 1 = 2 - 2r^2 \]
Rearranging gives:
\[ 2r^2 = 1 \quad \Rightarrow \quad r^2 = \frac{1}{2} \]
Thus:
\[ r = \pm \frac{1}{\sqrt{2}} \]

### Step 6: Determine the Correct Sign for \( r \)
Given that \( b_2 < 0 \) implies \( ar < 0 \). Since \( a > 0 \) (as it is the first term of the sequence), we must have:
\[ r < 0 \quad \Rightarrow \quad r = -\frac{1}{\sqrt{2}} \]

### Step 7: Substitute \( r \) Back to Find \( a \)
Now substituting \( r \) back into the expression for \( a \):
\[ a = \frac{1}{1 - \frac{1}{\sqrt{2}}} \]
To simplify:
\[ a = \frac{1}{\frac{\sqrt{2} - 1}{\sqrt{2}}} = \frac{\sqrt{2}}{\sqrt{2} - 1} \]

### Step 8: Rationalize the Denominator
Multiply the numerator and denominator by \( \sqrt{2} + 1 \):
\[ a = \frac{\sqrt{2}(\sqrt{2} + 1)}{(\sqrt{2} - 1)(\sqrt{2} + 1)} = \frac{2 + \sqrt{2}}{2 - 1} = 2 + \sqrt{2} \]

### Conclusion
Thus, the value of \( b_1 \) is:
\[ \boxed{2 + \sqrt{2}} \]