Key Concepts and Formulas

1. Sum of an Infinite Geometric Series (IGS): An infinite geometric series with first term $a$ and common ratio $r$ converges if and only if $|r| < 1$. The sum $S$ is given by $S = \frac{a}{1-r}$.
2. Series of Cubes of Terms: If the terms of an IGS are $a, ar, ar^2, \dots$, then the series of cubes of its terms is $a^3, (ar)^3, (ar^2)^3, \dots$, which simplifies to $a^3, a^3r^3, a^3r^6, \dots$. This is also an IGS with first term $a^3$ and common ratio $r^3$. For this series to converge, $|r^3| < 1$, which implies $|r| < 1$. The sum of this series is $S_c = \frac{a^3}{1-r^3}$.
3. Algebraic Identity: The difference of cubes formula is $x^3 - y^3 = (x-y)(x^2+xy+y^2)$. This will be useful for factoring $1-r^3$.
4. Positive Terms Condition: The problem states that the terms are positive. If the first term $a$ is positive, then for all terms to be positive, the common ratio $r$ must also be positive. Thus, for this problem, we require $0 < r < 1$.

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Step-by-Step Solution

Let the first term of the infinite geometric series be $a$ and the common ratio be $r$.

Step 1: Formulate equations from the given information.

We are given that the sum of the infinite geometric series is 3. Using the formula $S = \frac{a}{1-r}$: $\frac{a}{1-r} = 3 \quad \text{(Equation 1)}$ We are also given that the sum of the cubes of its terms is $\frac{27}{19}$. The series of cubes has the first term $a^3$ and common ratio $r^3$. Using the formula $S_c = \frac{\text{first term}}{1 - \text{common ratio}}$: $\frac{a^3}{1-r^3} = \frac{27}{19} \quad \text{(Equation 2)}$

Step 2: Manipulate Equation 2 to incorporate Equation 1.

We can rewrite Equation 2 by factoring the denominator $1-r^3$ using the difference of cubes identity: $\frac{a^3}{(1-r)(1+r+r^2)} = \frac{27}{19}$ Now, we can group terms to isolate the expression from Equation 1: $\left(\frac{a}{1-r}\right) \cdot \frac{a^2}{1+r+r^2} = \frac{27}{19}$

Step 3: Substitute the value from Equation 1 into the manipulated Equation 2.

Substitute $\frac{a}{1-r} = 3$ into the equation from Step 2: $3 \cdot \frac{a^2}{1+r+r^2} = \frac{27}{19}$ Divide both sides by 3: $\frac{a^2}{1+r+r^2} = \frac{9}{19} \quad \text{(Equation 3)}$

Step 4: Express $a^2$ in terms of $r$ and substitute into Equation 3.

From Equation 1, we can express $a$ in terms of $r$: $a = 3(1-r)$ Squaring both sides gives $a^2$: $a^2 = (3(1-r))^2 = 9(1-r)^2$ Now substitute this expression for $a^2$ into Equation 3: $\frac{9(1-r)^2}{1+r+r^2} = \frac{9}{19}$

Step 5: Solve the resulting equation for $r$.

Divide both sides by 9: $\frac{(1-r)^2}{1+r+r^2} = \frac{1}{19}$ Expand $(1-r)^2$: $\frac{1 - 2r + r^2}{1+r+r^2} = \frac{1}{19}$ Cross-multiply: $19(1 - 2r + r^2) = 1(1+r+r^2)$ $19 - 38r + 19r^2 = 1 + r + r^2$ Rearrange into a quadratic equation: $19r^2 - r^2 - 38r - r + 19 - 1 = 0$ $18r^2 - 39r + 18 = 0$ Divide the equation by 3 to simplify: $6r^2 - 13r + 6 = 0$ Factor the quadratic equation. We look for two numbers that multiply to $6 \times 6 = 36$ and add to $-13$. These numbers are $-4$ and $-9$. $6r^2 - 4r - 9r + 6 = 0$ $2r(3r - 2) - 3(3r - 2) = 0$ $(2r - 3)(3r - 2) = 0$ This gives two possible values for $r$: $2r - 3 = 0 \implies r = \frac{3}{2}$ $3r - 2 = 0 \implies r = \frac{2}{3}$

Step 6: Check the validity of the common ratio $r$.

From the problem statement, the terms of the series are positive. This implies that the first term $a$ is positive and the common ratio $r$ must also be positive. Furthermore, for an infinite geometric series to converge, the absolute value of the common ratio must be less than 1, i.e., $|r| < 1$. Therefore, we must have $0 < r < 1$.

Let's check our two potential values for $r$:

• If $r = \frac{3}{2}$, then $r > 1$. This violates the condition for convergence ($|r| < 1$), so this solution is extraneous.
• If $r = \frac{2}{3}$, then $0 < \frac{2}{3} < 1$. This satisfies the condition for convergence and the positivity requirement.

Thus, the common ratio of the series is $\frac{2}{3}$.

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Common Mistakes & Tips

• Convergence Condition: Always verify that the calculated common ratio $r$ satisfies $|r|<1$. An extraneous solution may arise from the algebraic manipulation.
• Positivity Constraint: Ensure the common ratio $r$ is positive if the problem states the terms are positive. This helps in eliminating invalid solutions.
• Algebraic Manipulation: Be careful when expanding and rearranging equations, especially when dealing with quadratic terms. The difference of cubes identity is a key tool here.

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Summary

The problem involves an infinite geometric series where the sum of the series and the sum of the cubes of its terms are given. We set up two equations using the formula for the sum of an infinite geometric series. By cleverly manipulating these equations, particularly by factoring the denominator of the sum of cubes formula, and substituting the first equation into the second, we reduced the problem to solving a quadratic equation for the common ratio $r$. Finally, we checked the validity of the obtained values of $r$ against the conditions for convergence and positivity, leading to the unique valid common ratio.

The final answer is $\boxed{\text{C}}$.