The second term of a geometric series is $^{-2}/_3$ and its sum to infinity is $^3/_2$. Find its common ratio.

Analysis

To solve this problem, we need to use the standard formulas for a geometric series:

1. The $n^{th}$ term: $T_n = ar^{n-1}$
2. The sum to infinity: $S_\infty = \frac{a}{1 - r}$

Where:

• $a$ is the first term
• $r$ is the common ratio

Concept

We are given:

• The second term ($T_2$) = $ar = -\frac{2}{3}$
• The sum to infinity ($S_\infty$) = $\frac{a}{1 - r} = \frac{3}{2}$

Step-by-Step Solution

Step 1: Express $a$ in terms of $r$ from the second term equation. From $ar = -\frac{2}{3}$, we get: $a = \frac{-2}{3r}$

Step 2: Substitute $a$ into the sum to infinity formula. $\frac{\frac{-2}{3r}}{1 - r} = \frac{3}{2}$

Simplify the left side: $\frac{-2}{3r(1 - r)} = \frac{3}{2}$

Step 3: Cross-multiply to solve for $r$. $3 \times [3r(1 - r)] = -2 \times 2$ $9r(1 - r) = -4$ $9r - 9r^2 = -4$

Rearrange into a standard quadratic equation ($ax^2 + bx + c = 0$): $9r^2 - 9r - 4 = 0$

Step 4: Solve the quadratic equation. Using the factorization method, we look for two numbers that multiply to $(9 \times -4) = -36$ and add to $-9$. These numbers are $-12$ and $3$. $9r^2 - 12r + 3r - 4 = 0$ $3r(3r - 4) + 1(3r - 4) = 0$ $(3r + 1)(3r - 4) = 0$

This gives us two possible values for $r$:

1. $3r + 1 = 0 \implies r = -\frac{1}{3}$
2. $3r - 4 = 0 \implies r = \frac{4}{3}$

Step 5: Evaluate the validity of the roots. For a geometric series to have a sum to infinity, the common ratio must satisfy $|r| < 1$.

• If $r = -\frac{1}{3}$, $|-\frac{1}{3}| < 1$ (Valid)
• If $r = \frac{4}{3}$, $|\frac{4}{3}| > 1$ (Invalid, as the series would diverge)

Therefore, the common ratio must be $-\frac{1}{3}$.

Conclusion

Comparing our result to the given options, the correct choice is A.

Exam Tip

Always remember the condition for the sum to infinity: $|r| < 1$. If you calculate a common ratio greater than 1 or less than -1, it cannot be the sum to infinity for that series. This quick check can help you eliminate wrong options immediately!