An infinite `G.P.` has `2^(nd)` term `x` and its sum is `4`. Then `x` belongs to

Options

A. `(0,2]`
B. `(1,8)`
C. `(-8,1]`
D. none of these

Explanation

To solve the problem step by step, we need to analyze the given information about the infinite geometric progression (G.P.) and derive the range for \( x \).

### Step 1: Understand the properties of the infinite G.P.
The sum \( S \) of an infinite G.P. is given by the formula:
\[ S = \frac{a}{1 - r} \]
where \( a \) is the first term and \( r \) is the common ratio. For the infinite G.P. to converge, \( |r| < 1 \).

### Step 2: Identify the terms of the G.P.
We know that the second term of the G.P. is \( x \). The first term \( a \) can be expressed in terms of \( x \) and \( r \):
\[ \text{Second term} = ar = x \implies a = \frac{x}{r} \]

### Step 3: Set up the equation for the sum
Given that the sum of the G.P. is 4, we can substitute \( a \) into the sum formula:
\[ 4 = \frac{\frac{x}{r}}{1 - r} \]
Multiplying both sides by \( r(1 - r) \):
\[ 4r(1 - r) = x \]
This simplifies to:
\[ x = 4r - 4r^2 \]

### Step 4: Analyze the function \( x = 4r - 4r^2 \)
This is a quadratic function in terms of \( r \). To find the maximum and minimum values of \( x \), we can complete the square or use calculus.

### Step 5: Differentiate to find critical points
We differentiate \( x \) with respect to \( r \):
\[ \frac{dx}{dr} = 4 - 8r \]
Setting the derivative to zero to find critical points:
\[ 4 - 8r = 0 \implies r = \frac{1}{2} \]

### Step 6: Determine if it is a maximum or minimum
To confirm whether this critical point is a maximum or minimum, we can check the second derivative:
\[ \frac{d^2x}{dr^2} = -8 \]
Since the second derivative is negative, \( r = \frac{1}{2} \) is a maximum.

### Step 7: Calculate the maximum value of \( x \)
Substituting \( r = \frac{1}{2} \) back into the equation for \( x \):
\[ x = 4\left(\frac{1}{2}\right) - 4\left(\frac{1}{2}\right)^2 = 2 - 1 = 1 \]

### Step 8: Determine the minimum value of \( x \)
Now we need to check the endpoints of the interval \( r \in (-1, 1) \):
- At \( r = -1 \):
\[ x = 4(-1) - 4(-1)^2 = -4 - 4 = -8 \]
- At \( r = 1 \):
\[ x = 4(1) - 4(1)^2 = 4 - 4 = 0 \]

### Step 9: Compile the range of \( x \)
Thus, the range of \( x \) is:
\[ [-8, 1] \]

### Final Answer
The value of \( x \) belongs to the interval:
\[ [-8, 1] \]