If the sum to infinty of the series , `1+4x+7x^(2)+10x^(3)+….`, is `(35)/(16)`, where `|x| lt 1`, then `'x'` equals to

Options

A. `19//7`
B. `1//5`
C. `1//4`
D. None of these

Explanation

To solve the problem, we need to find the value of \( x \) such that the sum to infinity of the series \( 1 + 4x + 7x^2 + 10x^3 + \ldots \) equals \( \frac{35}{16} \), where \( |x| < 1 \).

### Step-by-Step Solution:

1. **Identify the Series**:
The given series is \( 1 + 4x + 7x^2 + 10x^3 + \ldots \). The coefficients \( 1, 4, 7, 10 \) form an arithmetic progression (AP) with the first term \( a = 1 \) and common difference \( d = 3 \).

2. **General Term of the Series**:
The \( n \)-th term of the series can be expressed as:
\[ T_n = (1 + 3(n-1))x^{n-1} = (3n - 2)x^{n-1} \]

3. **Sum of the Series**:
The sum \( S \) of the series can be written as:
\[ S = \sum_{n=1}^{\infty} (3n - 2)x^{n-1} \]
This can be split into two separate sums:
\[ S = 3\sum_{n=1}^{\infty} nx^{n-1} - 2\sum_{n=1}^{\infty} x^{n-1} \]

4. **Using the Formula for the Sums**:
The sum \( \sum_{n=1}^{\infty} nx^{n-1} \) is given by:
\[ \sum_{n=1}^{\infty} nx^{n-1} = \frac{1}{(1-x)^2} \]
The sum \( \sum_{n=1}^{\infty} x^{n-1} \) is:
\[ \sum_{n=1}^{\infty} x^{n-1} = \frac{1}{1-x} \]

5. **Substituting the Sums**:
Substituting these results into the expression for \( S \):
\[ S = 3 \cdot \frac{1}{(1-x)^2} - 2 \cdot \frac{1}{1-x} \]

6. **Finding a Common Denominator**:
The common denominator is \( (1-x)^2 \):
\[ S = \frac{3}{(1-x)^2} - \frac{2(1-x)}{(1-x)^2} = \frac{3 - 2(1-x)}{(1-x)^2} \]
Simplifying the numerator:
\[ S = \frac{3 - 2 + 2x}{(1-x)^2} = \frac{1 + 2x}{(1-x)^2} \]

7. **Setting the Sum Equal to \( \frac{35}{16} \)**:
We set the expression for \( S \) equal to \( \frac{35}{16} \):
\[ \frac{1 + 2x}{(1-x)^2} = \frac{35}{16} \]

8. **Cross-Multiplying**:
Cross-multiplying gives:
\[ 16(1 + 2x) = 35(1 - 2x + x^2) \]

9. **Expanding and Rearranging**:
Expanding both sides:
\[ 16 + 32x = 35 - 70x + 35x^2 \]
Rearranging gives:
\[ 35x^2 - 102x + 19 = 0 \]

10. **Using the Quadratic Formula**:
Using the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
\[ x = \frac{102 \pm \sqrt{(-102)^2 - 4 \cdot 35 \cdot 19}}{2 \cdot 35} \]
Calculating the discriminant:
\[ = \frac{102 \pm \sqrt{10404 - 2660}}{70} = \frac{102 \pm \sqrt{7744}}{70} = \frac{102 \pm 88}{70} \]
Thus, we have two potential solutions:
\[ x = \frac{190}{70} = \frac{19}{7} \quad \text{and} \quad x = \frac{14}{70} = \frac{1}{5} \]

11. **Selecting the Valid Solution**:
Since \( |x| < 1 \), we discard \( \frac{19}{7} \) (as it is greater than 1) and accept:
\[ x = \frac{1}{5} \]

### Final Answer:
Thus, the value of \( x \) is \( \frac{1}{5} \).