If the sum to the infinity of the series `3+5r+7r^2+....` is `44/9` then find the value of r

Options

A. `(1)/(4)`
B. `(1)/(2)`
C. `(1)/(3)`
D. `(2)/(3)`

Answer

To solve the problem, we need to find the value of \( r \) given that the sum to infinity of the series \( 3 + 5r + 7r^2 + 9r^3 + \ldots \) is \( \frac{44}{9} \).

### Step-by-Step Solution:

1. **Define the Series**:
Let \( S \) be the sum of the series:
\[ S = 3 + 5r + 7r^2 + 9r^3 + \ldots \]

2. **Multiply by \( r \)**:
Multiply the entire series by \( r \):
\[ rS = 3r + 5r^2 + 7r^3 + 9r^4 + \ldots \]

3. **Shift the Series**:
Now, we can subtract \( rS \) from \( S \):
\[ S - rS = (3 + 5r + 7r^2 + 9r^3 + \ldots) - (3r + 5r^2 + 7r^3 + 9r^4 + \ldots) \]
This simplifies to:
\[ S(1 - r) = 3 + (5r - 3r) + (7r^2 - 5r^2) + (9r^3 - 7r^3) + \ldots \]
Which results in:
\[ S(1 - r) = 3 + 2r + 2r^2 + 2r^3 + \ldots \]

4. **Recognize the Infinite Geometric Series**:
The series \( 2r + 2r^2 + 2r^3 + \ldots \) can be factored out:
\[ S(1 - r) = 3 + 2r(1 + r + r^2 + \ldots) \]
The sum of the infinite geometric series \( 1 + r + r^2 + \ldots \) is given by \( \frac{1}{1 - r} \) (valid for \( |r| < 1 \)):
\[ S(1 - r) = 3 + 2r \cdot \frac{1}{1 - r} \]

5. **Substituting the Value of \( S \)**:
We know \( S = \frac{44}{9} \):
\[ \frac{44}{9}(1 - r) = 3 + \frac{2r}{1 - r} \]

6. **Multiply Through by \( 9(1 - r) \)**:
To eliminate the fraction, multiply both sides by \( 9(1 - r) \):
\[ 44(1 - r) = 27(1 - r) + 18r \]
Simplifying gives:
\[ 44 - 44r = 27 - 27r + 18r \]
Which simplifies to:
\[ 44 - 44r = 27 - 9r \]

7. **Rearranging the Equation**:
Move all terms involving \( r \) to one side:
\[ 44 - 27 = 44r - 9r \]
This simplifies to:
\[ 17 = 35r \]

8. **Solving for \( r \)**:
Finally, divide both sides by 35:
\[ r = \frac{17}{35} \]

### Final Answer:
The value of \( r \) is \( \frac{17}{35} \).