The sum of an infinite geometric series with positive terms is 3 and the sums of the cubes of its terms is `(27)/(19)`. Then the common ratio of this series is

Options

A. `(4)/(9)`
B. `(2)/(9)`
C. `(2)/(3)`
D. `(1)/(3)`

Answer

To solve the problem, we need to find the common ratio \( r \) of an infinite geometric series given the sum of the series and the sum of the cubes of its terms.

### Step-by-Step Solution:

1. **Define the Series**:
Let the first term of the geometric series be \( a \) and the common ratio be \( r \). The terms of the series are \( a, ar, ar^2, ar^3, \ldots \).

2. **Sum of the Infinite Series**:
The sum \( S \) of an infinite geometric series can be expressed as:
\[ S = \frac{a}{1 - r} \]
Given that the sum is equal to 3, we have:
\[ \frac{a}{1 - r} = 3 \quad \text{(1)} \]

3. **Sum of the Cubes of the Terms**:
The sum of the cubes of the terms is:
\[ a^3 + (ar)^3 + (ar^2)^3 + \ldots = a^3 + a^3 r^3 + a^3 r^6 + \ldots \]
This can be simplified using the formula for the sum of an infinite geometric series:
\[ \text{Sum of cubes} = \frac{a^3}{1 - r^3} \]
Given that this sum is equal to \( \frac{27}{19} \), we have:
\[ \frac{a^3}{1 - r^3} = \frac{27}{19} \quad \text{(2)} \]

4. **Substituting \( a \) from Equation (1)**:
From equation (1), we can express \( a \) in terms of \( r \):
\[ a = 3(1 - r) \]
Substitute this value of \( a \) into equation (2):
\[ \frac{(3(1 - r))^3}{1 - r^3} = \frac{27}{19} \]
Simplifying the left side:
\[ \frac{27(1 - r)^3}{1 - r^3} = \frac{27}{19} \]

5. **Cancelling \( 27 \)**:
We can cancel \( 27 \) from both sides:
\[ \frac{(1 - r)^3}{1 - r^3} = \frac{1}{19} \]

6. **Cross Multiplying**:
Cross-multiplying gives:
\[ 19(1 - r)^3 = 1 - r^3 \]

7. **Expanding and Rearranging**:
Expanding \( (1 - r)^3 \):
\[ (1 - r)^3 = 1 - 3r + 3r^2 - r^3 \]
Thus:
\[ 19(1 - 3r + 3r^2 - r^3) = 1 - r^3 \]
Expanding this:
\[ 19 - 57r + 57r^2 - 19r^3 = 1 - r^3 \]
Rearranging gives:
\[ 18r^3 - 57r^2 + 57r + 18 = 0 \]

8. **Dividing by 3**:
Dividing the entire equation by 3:
\[ 6r^3 - 19r^2 + 19r + 6 = 0 \]

9. **Factoring the Cubic Equation**:
We can factor this cubic equation:
\[ (3r - 2)(2r - 3) = 0 \]
This gives us:
\[ r = \frac{2}{3} \quad \text{or} \quad r = \frac{3}{2} \]

10. **Selecting the Valid Solution**:
Since \( r \) must be between 0 and 1, we reject \( r = \frac{3}{2} \). Therefore, the common ratio is:
\[ r = \frac{2}{3} \]

### Final Answer:
The common ratio \( r \) of the series is \( \frac{2}{3} \).