College Algebra Quiz: Sum Of Geometric Series

What this quiz covers

This quiz focuses on Sum Of Geometric Series, giving you a quick way to practice the rules, question types, and explanations that matter most for College Algebra.

How to use this quiz

Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

Question 1

1. Converges for $-4 < x < 2$; sum is $\frac{3(2x-1)}{2-x}$ (correct answer)
2. Converges for $-4 < x < 2$; sum is $\frac{6(2x-1)}{2-x}$
3. Converges for $-2 < x < 4$; sum is $\frac{3(2x-1)}{2-x}$
4. Converges for $-4 < x < 2$; sum is $\frac{9(2x-1)}{2-x}$

Explanation: This is an infinite geometric series with first term $a = 2x-1$ and common ratio $r = \frac{x+1}{3}$. For convergence, we need $|r| < 1$, which means $\left|\frac{x+1}{3}\right| < 1$. This gives $|x+1| < 3$, so $-3 < x+1 < 3$, which means $-4 < x < 2$. When the series converges, its sum is $S = \frac{a}{1-r} = \frac{2x-1}{1-\frac{x+1}{3}} = \frac{2x-1}{\frac{3-(x+1)}{3}} = \frac{2x-1}{\frac{2-x}{3}} = \frac{3(2x-1)}{2-x}$.

Question 2

1. $r = \frac{1}{4}$ (correct answer)
2. $r = \frac{1}{3}$
3. $r = \frac{3}{4}$
4. $r = \frac{2}{3}$

Explanation: For an infinite geometric series with $|r| < 1$, the sum is $S = \frac{a}{1-r}$. Substituting the given values: $24 = \frac{18}{1-r}$. Solving for $r$: $24(1-r) = 18$, so $24 - 24r = 18$, which gives $24r = 6$, and therefore $r = \frac{1}{4}$. We can verify: $|\frac{1}{4}| < 1$ (convergence condition satisfied) and $S = \frac{18}{1-\frac{1}{4}} = \frac{18}{\frac{3}{4}} = 24$ ✓. Choice B gives $r = \frac{1}{3}$: $S = \frac{18}{1-\frac{1}{3}} = \frac{18}{\frac{2}{3}} = 27 \neq 24$. Choice C gives $r = \frac{3}{4}$: $S = \frac{18}{1-\frac{3}{4}} = \frac{18}{\frac{1}{4}} = 72 \neq 24$. Choice D gives $r = \frac{2}{3}$: $S = \frac{18}{1-\frac{2}{3}} = \frac{18}{\frac{1}{3}} = 54 \neq 24$.

Question 3

1. $S = 189$ (correct answer)
2. $S = 192$
3. $S = 195$
4. $S = 198$

Explanation: Let's rewrite this series by substituting $j = k - 2$, so when $k = 2$, $j = 0$ and when $k = 7$, $j = 5$. The series becomes $\sum_{j=0}^{5} 3 \cdot 2^j$. This is a geometric series with first term $a = 3 \cdot 2^0 = 3$, common ratio $r = 2$, and 6 terms (from $j = 0$ to $j = 5$). Using $S_n = a \cdot \frac{r^n - 1}{r - 1}$: $S_6 = 3 \cdot \frac{2^6 - 1}{2 - 1} = 3 \cdot \frac{64 - 1}{1} = 3 \cdot 63 = 189$. Alternatively, we can compute directly: the terms are $3 \cdot 2^0 = 3$, $3 \cdot 2^1 = 6$, $3 \cdot 2^2 = 12$, $3 \cdot 2^3 = 24$, $3 \cdot 2^4 = 48$, $3 \cdot 2^5 = 96$. Sum: $3 + 6 + 12 + 24 + 48 + 96 = 189$. Choice B (192) might result from including an extra term or a calculation error. Choice C (195) and D (198) are further off and likely represent other common errors.

Question 4

1. $S_6 = 12600$ (correct answer)
2. $S_6 = 12700$
3. $S_6 = 12800$
4. $S_6 = 12900$

Explanation: The geometric sequence has first term $a = 200$ and common ratio $r = 2$ (since the population doubles). Using the formula $S_n = a \cdot \frac{r^n - 1}{r - 1}$: $S_6 = 200 \cdot \frac{2^6 - 1}{2 - 1} = 200 \cdot \frac{64 - 1}{1} = 200 \cdot 63 = 12600$. We can verify: the first 6 terms are 200, 400, 800, 1600, 3200, 6400, and $200 + 400 + 800 + 1600 + 3200 + 6400 = 12600$.