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

Options

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

Explanation

To solve the problem, we need to find the common ratio 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. **Understanding the Infinite Geometric Series**:
The sum \( S \) of an infinite geometric series with first term \( a \) and common ratio \( r \) (where \( |r| < 1 \)) is given by:
\[ S = \frac{a}{1 - r} \]
Given that the sum is 3, we have:
\[ \frac{a}{1 - r} = 3 \quad \text{(1)} \]

2. **Sum of the Cubes of the Terms**:
The terms of the series are \( a, ar, ar^2, ar^3, \ldots \). The cubes of these terms are \( a^3, (ar)^3, (ar^2)^3, (ar^3)^3, \ldots \), which form another geometric series with first term \( a^3 \) and common ratio \( r^3 \).
The sum of the cubes \( S_c \) is given by:
\[ S_c = \frac{a^3}{1 - r^3} \]
Given that the sum of the cubes is \( \frac{27}{19} \), we have:
\[ \frac{a^3}{1 - r^3} = \frac{27}{19} \quad \text{(2)} \]

3. **Expressing \( a \) in terms of \( r \)**:
From equation (1), we can express \( a \) as:
\[ a = 3(1 - r) \quad \text{(3)} \]

4. **Substituting \( a \) into the Sum of Cubes**:
Substitute equation (3) 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. **Cross-Multiplying**:
Cross-multiply to eliminate the fraction:
\[ 27(1 - r)^3 \cdot 19 = 27(1 - r^3) \]
Dividing both sides by 27:
\[ 19(1 - r)^3 = 1 - r^3 \quad \text{(4)} \]

6. **Expanding and Rearranging**:
Expand the left side:
\[ 19(1 - 3r + 3r^2 - r^3) = 1 - r^3 \]
This simplifies to:
\[ 19 - 57r + 57r^2 - 19r^3 = 1 - r^3 \]
Rearranging gives:
\[ 18 = 57r - 57r^2 + 18r^3 \]
Rearranging further:
\[ 18r^3 - 57r^2 + 57r - 18 = 0 \quad \text{(5)} \]

7. **Finding the Roots**:
We can use the Rational Root Theorem or synthetic division to find the roots of the cubic equation (5). Testing \( r = \frac{1}{3} \):
\[ 18\left(\frac{1}{3}\right)^3 - 57\left(\frac{1}{3}\right)^2 + 57\left(\frac{1}{3}\right) - 18 = 0 \]
This simplifies to:
\[ 18 \cdot \frac{1}{27} - 57 \cdot \frac{1}{9} + 19 - 18 = 0 \]
Thus, \( r = \frac{1}{3} \) is a root.

8. **Conclusion**:
The common ratio \( r \) of the series is:
\[ r = \frac{1}{3} \]