Sum of an infinite G.P is 2 and sum of its two terms is 1.If its second terms is negative then which of the following is /are true ?

Options

A. one of the possible values of the first terms is `(2-sqrt(2))`
B. one of the possible vlaues of the first terms is `(2+sqrt(2))`
C. one of the possible values of the common ratio is `(sqrt(2)-1)`
D. one of the possible values of the common ratio is `(1)/(sqrt(2))`

Explanation

To solve the problem step by step, we need to analyze the information given about the infinite geometric progression (G.P.).

### Step 1: Write down the formulas for the sum of an infinite G.P. and the sum of the first two terms.

The sum \( S \) of an infinite G.P. with first term \( a \) and common ratio \( r \) (where \( |r| < 1 \)) is given by:

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

We are given that the sum of the infinite G.P. is 2:

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

The sum of the first two terms of the G.P. is:

\[ S_2 = a + ar = a(1 + r) \]

We are also given that the sum of the first two terms is 1:

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

### Step 2: Solve the equations simultaneously.

From equation (1):

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

Substituting equation (3) into equation (2):

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

Expanding this:

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

Dividing both sides by 2:

\[ 1 - r^2 = \frac{1}{2} \]

Rearranging gives:

\[ r^2 = 1 - \frac{1}{2} = \frac{1}{2} \]

Taking the square root of both sides:

\[ r = \pm \frac{1}{\sqrt{2}} \]

### Step 3: Find the corresponding values of \( a \).

Using \( r = -\frac{1}{\sqrt{2}} \) (since the second term is negative):

Substituting \( r \) back into equation (3):

\[ a = 2\left(1 - \left(-\frac{1}{\sqrt{2}}\right)\right) = 2\left(1 + \frac{1}{\sqrt{2}}\right) = 2 + \frac{2}{\sqrt{2}} = 2 + \sqrt{2} \]

Now, using \( r = \frac{1}{\sqrt{2}} \):

\[ a = 2\left(1 - \frac{1}{\sqrt{2}}\right) = 2 - \frac{2}{\sqrt{2}} = 2 - \sqrt{2} \]

### Step 4: Conclusion

The possible values for \( a \) are \( 2 + \sqrt{2} \) and \( 2 - \sqrt{2} \). Since we are looking for the case where the second term is negative, we have:

1. \( r = -\frac{1}{\sqrt{2}} \) gives \( a = 2 + \sqrt{2} \)
2. \( r = \frac{1}{\sqrt{2}} \) gives \( a = 2 - \sqrt{2} \)

Thus, the options that are true are those corresponding to these values of \( a \).