AP Calculus BC Quiz: Working With Geometric Series

What this quiz covers

This quiz focuses on Working With Geometric Series, giving you a quick way to practice the rules, question types, and explanations that matter most for AP Calculus BC.

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. $6.25$ (correct answer)
2. $4$
3. $-6.25$
4. $16$
5. Diverges

Explanation: This problem involves finding the sum of an infinite geometric series of light intensities. A geometric series converges if the absolute value of the common ratio |r| is less than 1. Here, the first term a = 10 and r = -0.6, so |r| = 0.6 < 1, ensuring convergence. The sum is S = a / (1 - r) = 10 / (1 - (-0.6)) = 10 / 1.6 = 6.25. One tempting distractor is -6.25, which might result from using 1 + r instead of 1 - r in the denominator. When working with geometric series in iterative processes like reflections, verify the sign of the ratio to apply the sum formula correctly.

Question 2

1. $-\tfrac{2}{3}$
2. $\tfrac{2}{3}$
3. $\tfrac{5}{3}$
4. $\tfrac{7}{3}$
5. Diverges (correct answer)

Explanation: This problem involves determining if an infinite geometric series converges and finding its sum if it does. A geometric series converges if the absolute value of the common ratio |r| is less than 1. Here, the first term a = 1 and r = 5/2, so |r| = 5/2 > 1, meaning the series diverges. There is no finite sum since the terms grow without bound. One tempting distractor is -2/3, which might come from incorrectly applying the sum formula despite divergence, perhaps ignoring the condition on |r|. When evaluating geometric series, always check the convergence condition before attempting to compute the sum.

Question 3

1. Converges to $\tfrac{12}{1-1.05}$
2. Converges to $\tfrac{12}{1-0.05}$
3. Diverges because $|r|>1$ (correct answer)
4. Converges to $\tfrac{1.05}{1-12}$
5. Diverges because $|r|<1$

Explanation: This problem involves working with geometric series to determine if the total light intensity converges. A geometric series converges if the absolute value of the common ratio |r| is less than 1. Here, the intensities form a series with first term 12 and r = 1.05, but |1.05| > 1, so the series diverges. The sum formula S = a / (1 - r) only applies when |r| < 1, which is not the case here. One tempting distractor is converging to 12/(1-1.05), which fails because the condition |r| < 1 is not met, making the formula invalid. A transferable strategy for geometric series is to identify the first term and common ratio, check the convergence condition |r| < 1, and only then apply the sum formula if appropriate.

Question 4

1. $\tfrac{3}{8}\cdot\tfrac{1}{1-\tfrac{2}{3}}$ (correct answer)
2. $\tfrac{3}{8}\cdot\tfrac{1}{1-\tfrac{3}{2}}$
3. $\tfrac{3}{8}\cdot\tfrac{1}{1+\tfrac{2}{3}}$
4. Diverges because $\tfrac{2}{3}$ is not an integer
5. $\tfrac{3}{8}\left(\tfrac{2}{3}\right)\cdot\tfrac{1}{1-\tfrac{2}{3}}$

Explanation: This problem asks for the sum of a straightforward geometric series. The series ∑(n=0 to ∞) (3/8)(2/3)ⁿ has first term a = 3/8 (when n=0) and common ratio r = 2/3. Since |r| = 2/3 < 1, the series converges to a/(1-r) = (3/8)/(1-2/3) = (3/8)/(1/3) = (3/8)·3 = 9/8. This can also be written as (3/8)·(1/(1-2/3)) as shown in choice A. Choice B incorrectly uses 3/2 as the ratio (the reciprocal), while choice E unnecessarily includes an extra factor of 2/3. For geometric series of the form c·rⁿ starting at n=0, the sum is simply c/(1-r) when |r| < 1.

Question 5

1. Converges to $\tfrac{1}{1-\tfrac{5}{2}}$
2. Converges to $\tfrac{1}{1-\tfrac{2}{5}}$
3. Diverges because $\left|\tfrac{5}{2}\right|>1$ (correct answer)
4. Diverges because $\left|\tfrac{5}{2}\right|<1$
5. Converges to $\tfrac{1}{1+\tfrac{5}{2}}$

Explanation: This problem tests understanding of the convergence condition for geometric series. The series ∑(n=0 to ∞) (5/2)ⁿ has first term a = 1 and common ratio r = 5/2. For a geometric series to converge, we need |r| < 1. Here, |r| = |5/2| = 5/2 = 2.5 > 1, so the series diverges. The terms (5/2)ⁿ grow without bound as n increases, preventing the series from having a finite sum. Choice A incorrectly attempts to apply the sum formula despite divergence, while choice D gives the wrong reason for divergence. Remember that geometric series converge if and only if |r| < 1; when |r| ≥ 1, the series diverges regardless of the sign of r.

Question 6

1. $4500$
2. $5000$
3. Diverges
4. $\tfrac{500}{0.9}$
5. $\tfrac{500}{1-0.9}$ (correct answer)

Explanation: This problem requires working with geometric series to find the total amount deposited in a savings account. A geometric series converges if the absolute value of the common ratio $|r|$ is less than 1. Here, the deposits start with $500$ and continue with $r = 0.9$, and since $|0.9| < 1$, the series converges. The sum is $S = 500 / (1 - 0.9) = 500 / 0.1 = 5000$. One tempting distractor is 'diverges,' which fails because $|r| = 0.9 < 1$, so the series does converge to a finite value. A transferable strategy for geometric series is to identify the first term and common ratio, confirm $|r| < 1$ for convergence, and apply the sum formula for infinite terms.

Question 7

1. $48$
2. $72$ (correct answer)
3. $40$
4. $64$
5. Diverges

Explanation: This problem requires working with geometric series to find the total distance traveled by a bouncing ball. A geometric series converges if the absolute value of the common ratio |r| is less than 1. Here, the series is formed by the initial travel of 24 ft followed by subsequent travels scaled by r = 2/3 each time. Since |2/3| < 1, the series converges, and the sum is given by the formula S = a / (1 - r), where a = 24, yielding S = 24 / (1/3) = 72. One tempting distractor is 48, which might arise from incorrectly doubling only part of the series or misidentifying the first term. A transferable strategy for geometric series is to identify the first term and common ratio, confirm |r| < 1 for convergence, and apply the sum formula while considering the physical context.

Question 8

1. $\tfrac{4}{3}$
2. $2$ (correct answer)
3. $4$
4. $\tfrac{8}{3}$
5. Diverges

Explanation: This problem involves finding the sum of an infinite geometric series. A geometric series converges if the absolute value of the common ratio $|r|$ is less than 1. Here, the series is $4 \times \sum_{n=1}^{\infty} \left(\frac{1}{3}\right)^n$, with $r = \frac{1}{3} < 1$, ensuring convergence. The sum is $4 \times \frac{\frac{1}{3}}{1 - \frac{1}{3}} = 4 \times \frac{\frac{1}{3}}{\frac{2}{3}} = 4 \times \frac{1}{2} = 2$. One tempting distractor is 4, which could come from mistakenly including the n=0 term or misapplying the formula. When summing geometric series starting from n=1, remember to use the formula for the tail sum appropriately.

Question 9

1. Converges to $\tfrac{9}{1-(-2)}$
2. Converges to $\tfrac{9}{1-2}$
3. Diverges because $|r|>1$ (correct answer)
4. Converges to $\tfrac{-18}{1-(-2)}$
5. Diverges because $|r|<1$

Explanation: This problem involves working with geometric series to determine if the sum of areas converges. A geometric series converges if the absolute value of the common ratio |r| is less than 1. Here, the areas are 9, -18, 36, -72, ..., with r = -2, but |-2| > 1, so the series diverges. The sum formula S = a / (1 - r) only applies when |r| < 1, which is not satisfied here. One tempting distractor is converging to 9/(1-(-2)), which fails because |r| > 1 violates the convergence condition, making the sum infinite. A transferable strategy for geometric series is to identify the first term and common ratio, check the convergence condition |r| < 1 first, and only apply the sum formula if the series converges.

Question 10

1. $\dfrac{256}{3}$ (correct answer)
2. $84$
3. $\dfrac{64}{3}$
4. $\dfrac{80}{3}$
5. Diverges

Explanation: This problem involves finding the sum of an infinite geometric series of square areas. The series is 64 + 16 + 4 + ..., where the first term a = 64 and the common ratio r = 16/64 = 1/4 = 0.25. Since |r| = 0.25 < 1, the series converges. Applying the sum formula S = a/(1-r), we get S = 64/(1-0.25) = 64/0.75 = 64/(3/4) = 256/3. Choice C (64/3) might result from forgetting the first term or making an arithmetic error. The strategy for geometric series is to identify the pattern, verify convergence with |r| < 1, then carefully apply the sum formula S = a/(1-r).

Question 11

1. $21$
2. $\dfrac{21}{7}$
3. $\dfrac{7}{1-\frac{4}{3}}$
4. Diverges (correct answer)
5. $\dfrac{7}{1-\frac{3}{4}}$

Explanation: This problem asks whether the geometric series Σ(n=0 to ∞) 7(4/3)^n converges. In this series, the first term is a = 7(4/3)^0 = 7 and the common ratio is r = 4/3. For a geometric series to converge, we need |r| < 1. Here, |4/3| = 4/3 ≈ 1.33 > 1, so the series diverges. Choice C shows the sum formula 7/(1-4/3), which would give a negative denominator, confirming divergence. Choice E (7/(1-3/4)) represents a different series with r = 3/4 that would converge. When analyzing geometric series, always check the convergence condition |r| < 1 before attempting to find a sum.

Question 12

1. $36$
2. $48$ (correct answer)
3. $24$
4. $\tfrac{12}{1-\tfrac{3}{4}}$
5. $\tfrac{12}{1+\tfrac{3}{4}}$

Explanation: This problem requires finding the sum of a geometric series representing the ball's total travel distance. The ball travels 12 m initially, then 12(3/4) m, then 12(3/4)² m, and so on, giving us the series 12 + 12(3/4) + 12(3/4)² + ... = 12[1 + 3/4 + (3/4)² + ...]. This is a geometric series with first term a = 12 and common ratio r = 3/4. Since |r| = 3/4 < 1, the series converges to a/(1-r) = 12/(1-3/4) = 12/(1/4) = 48. Choice D incorrectly uses only the formula without the initial term factored out, while choice E uses the wrong sign in the denominator. When working with geometric series in applications, always identify the first term and common ratio, then apply the formula a/(1-r) for |r| < 1.

Question 13

1. $\tfrac{6}{1-\tfrac{1}{2}}$
2. $\tfrac{6\left(\tfrac{1}{2}\right)^2}{1-\tfrac{1}{2}}$ (correct answer)
3. $\tfrac{6\left(\tfrac{1}{2}\right)}{1-\tfrac{1}{2}}$
4. Diverges because it starts at $n=2$
5. $\tfrac{6\left(\tfrac{1}{2}\right)^2}{1+\tfrac{1}{2}}$

Explanation: This problem involves a geometric series starting at n=2, requiring careful handling of the initial index. The series ∑(n=2 to ∞) 6(1/2)ⁿ equals 6(1/2)² + 6(1/2)³ + ... = 6(1/2)²[1 + 1/2 + (1/2)² + ...]. Factoring out 6(1/2)², we get a standard geometric series with first term a = 1 and ratio r = 1/2. Since |r| = 1/2 < 1, this inner series sums to 1/(1-1/2) = 2, giving us a total of 6(1/2)² · 2 = 6(1/4) · 2 = 3/2. Alternatively, we can compute directly as 6(1/2)²/(1-1/2) = (6/4)/(1/2) = 3/2. Choice A incorrectly treats it as starting from n=0, while choice D wrongly claims divergence. For series starting at n=k with k>0, factor out the common terms to reduce to a standard geometric series.

Question 14

1. $\dfrac{15}{4}$ (correct answer)
2. $\dfrac{5}{1-\frac{1}{3}}$
3. $\dfrac{5}{3}$
4. Diverges
5. $\dfrac{15}{2}$

Explanation: This problem requires finding the sum of the geometric series 5/2 + 5/6 + 5/18 + .... We can factor out 5 to get 5(1/2 + 1/6 + 1/18 + ...), where the series in parentheses has first term a = 1/2 and common ratio r = (1/6)/(1/2) = 1/3. Since |r| = 1/3 < 1, the series converges. The sum of the inner series is (1/2)/(1-1/3) = (1/2)/(2/3) = 3/4, so the total sum is 5(3/4) = 15/4. Choice B (5/(1-1/3)) represents the sum if the first term were 5, not 5/2. When dealing with factored geometric series, carefully identify what remains as the first term after factoring.

Question 15

1. $\tfrac{1000}{1-0.7}$ (correct answer)
2. $\tfrac{1000}{1-\tfrac{10}{7}}$
3. $1700$
4. $\tfrac{1000}{0.7}$
5. Diverges

Explanation: This problem involves working with geometric series to find the total payments in a sequence. A geometric series converges if the absolute value of the common ratio $|r|$ is less than 1. Here, the payments are 1000, 700, 490, ..., with r = 0.7, and since $|0.7|$ < 1, the series converges. The sum is $S = \frac{1000}{1 - 0.7} = \frac{1000}{0.3} \approx 3333.33$. One tempting distractor is $1000 / 0.7$, which fails because it divides by r instead of (1 - r), ignoring the formula structure. A transferable strategy for geometric series is to identify the first term and common ratio, confirm $|r|$ < 1 for convergence, and apply the sum formula precisely.

Question 16

1. $128$ (correct answer)
2. $\tfrac{80}{1-\tfrac{3}{8}}$
3. $\tfrac{80}{1-\tfrac{8}{3}}$
4. $140$
5. Diverges

Explanation: This problem involves working with geometric series to find the total frames rendered by a computer. A geometric series converges if the absolute value of the common ratio $|r|$ is less than 1. Here, the frames start with 80 and continue with $r = \tfrac{3}{8}$, and since $|\tfrac{3}{8}| < 1$, the series converges. The sum is $S = 80 / (1 - \tfrac{3}{8}) = 80 / (\tfrac{5}{8}) = 128$. One tempting distractor is $80/(1-\tfrac{8}{3})$, which fails because it reverses the ratio fraction, leading to a negative and incorrect sum. A transferable strategy for geometric series is to identify the first term and common ratio, confirm $|r| < 1$ for convergence, and apply the sum formula accurately.

Question 17

1. $1$
2. $2$ (correct answer)
3. $6$
4. $\tfrac{3}{1+\tfrac{1}{2}}$
5. Diverges

Explanation: This problem requires working with geometric series to find the total dye added by the artist. A geometric series converges if the absolute value of the common ratio |r| is less than 1. Here, the amounts start with 3 mL and continue with r = -1/2, and since |-1/2| < 1, the series converges. The sum is S = 3 / (1 - (-1/2)) = 3 / (3/2) = 2. One tempting distractor is 6, which might come from incorrectly using r = 1/2 without the negative sign, leading to 3 / (1/2) = 6. A transferable strategy for geometric series is to identify the first term and common ratio, confirm |r| < 1 for convergence, and apply the sum formula, paying attention to signs for alternating series.

Question 18

1. $96$
2. $84$ (correct answer)
3. $72$
4. $60$
5. Diverges

Explanation: This problem involves finding the sum of an infinite geometric series to determine the total distance a bouncing ball travels. A geometric series converges if the absolute value of the common ratio |r| is less than 1. Here, the common ratio r = 3/4, so |r| = 3/4 < 1, ensuring convergence. The total distance is the initial drop of 12 meters plus twice the sum of the infinite geometric series of bounce heights, which is 2 * (12 * (3/4) / (1 - 3/4)) = 2 * (9 / (1/4)) = 2 * 36 = 72, plus 12 equals 84 meters. One tempting distractor is 96, which arises from incorrectly doubling the entire sum including the initial drop, as if treating the initial drop as part of the repeated up-and-down motions. When dealing with geometric series in physical contexts like bouncing balls, always separate the initial term if it's not repeated in the same way as the subsequent terms.

Question 19

1. $\tfrac{4}{7}$ (correct answer)
2. $\tfrac{7}{4}$
3. $\tfrac{1}{7}$
4. $4$
5. Diverges

Explanation: This problem involves finding the sum of an infinite geometric series. A geometric series converges if the absolute value of the common ratio |r| is less than 1. Here, the first term a = 1 and r = -3/4, so |r| = 3/4 < 1, ensuring convergence. The sum is S = a / (1 - r) = 1 / (1 - (-3/4)) = 1 / (7/4) = 4/7. One tempting distractor is 7/4, which could come from inverting the fraction incorrectly or confusing the formula. When evaluating alternating geometric series, double-check the denominator to ensure it's 1 minus the ratio.

Question 20

1. $15$ (correct answer)
2. $\tfrac{15}{2}$
3. $10$
4. $\tfrac{5}{3}$
5. Diverges

Explanation: This problem involves finding the sum of an infinite geometric series. A geometric series converges if the absolute value of the common ratio |r| is less than 1. Here, the first term a = 5 and r = 2/3, so |r| = 2/3 < 1, ensuring convergence. The sum is given by S = a / (1 - r) = 5 / (1 - 2/3) = 5 / (1/3) = 15. One tempting distractor is 15/2, which might result from mistakenly using the formula for the sum starting from n=1 instead of n=0, excluding the initial term. When working with geometric series, always identify the first term and common ratio accurately to apply the sum formula correctly.