If x is the first term of a G.P. with infinite number of terms and `S_(oo) = 5`, then x is

Options

A. `0 lt x lt 10`
B. `x ge 10`
C. `x lt -10`
D. `-10 lt x lt 0`

Answer

To solve the problem, we need to find the value of \( x \) given that it is the first term of a geometric progression (G.P.) with an infinite number of terms and that the sum to infinity \( S_{\infty} = 5 \).

### Step-by-Step Solution:

1. **Understanding the Sum of Infinite G.P.:**
The formula for the sum of an infinite geometric progression is given by:
\[ S_{\infty} = \frac{a}{1 - r} \]
where \( a \) is the first term and \( r \) is the common ratio.

2. **Setting Up the Equation:**
In our case, the first term \( a = x \) and the sum \( S_{\infty} = 5 \). Thus, we can write:
\[ \frac{x}{1 - r} = 5 \]

3. **Rearranging the Equation:**
From the equation above, we can rearrange it to express \( x \) in terms of \( r \):
\[ x = 5(1 - r) = 5 - 5r \]

4. **Condition for the Common Ratio:**
For the sum of an infinite G.P. to exist, the absolute value of the common ratio must be less than 1:
\[ |r| < 1 \]

5. **Expressing the Condition:**
We can express \( r \) in terms of \( x \):
\[ r = \frac{5 - x}{5} \]
Now, substituting this into the condition \( |r| < 1 \):
\[ \left| \frac{5 - x}{5} \right| < 1 \]

6. **Solving the Inequality:**
This absolute value inequality can be split into two inequalities:
\[ -1 < \frac{5 - x}{5} < 1 \]

- For the right side:
\[ \frac{5 - x}{5} < 1 \implies 5 - x < 5 \implies -x < 0 \implies x > 0 \]

- For the left side:
\[ -1 < \frac{5 - x}{5} \implies -5 < 5 - x \implies -5 + 5 < -x \implies 0 < -x \implies x < 10 \]

7. **Combining the Results:**
From the inequalities, we find:
\[ 0 < x < 10 \]

### Conclusion:
Thus, the value of \( x \) must be greater than 0 and less than 10.