College Algebra Quiz: Sum Of Geometric Series
4 questions · exam conditions
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Sum Of Geometric SeriesQuestion 1 of 4

A certain bacteria culture doubles every 3 hours. If the culture starts with 200 bacteria, the population after 3n3n hours forms a geometric sequence: 200, 400, 800, 1600, .... What is the sum of the first 6 terms of this sequence?

S6=12600S_6 = 12600
S6=12700S_6 = 12700
S6=12800S_6 = 12800
S6=12900S_6 = 12900
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College Algebra Quiz

College Algebra Quiz: Sum Of Geometric Series

Practice Sum Of Geometric Series in College Algebra with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

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.

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Question 1

A certain bacteria culture doubles every 3 hours. If the culture starts with 200 bacteria, the population after 3n3n hours forms a geometric sequence: 200, 400, 800, 1600, .... What is the sum of the first 6 terms of this sequence?

  1. S6=12600S_6 = 12600 (correct answer)
  2. S6=12700S_6 = 12700
  3. S6=12800S_6 = 12800
  4. S6=12900S_6 = 12900
Explanation: The geometric sequence has first term a=200a = 200 and common ratio r=2r = 2 (since the population doubles). Using the formula Sn=a⋅rn−1r−1S_n = a \cdot \frac{r^n - 1}{r - 1}: S6=200⋅26−12−1=200⋅64−11=200⋅63=12600S_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=12600200 + 400 + 800 + 1600 + 3200 + 6400 = 12600.

Question 2

For what value of xx does the infinite geometric series ∑n=0∞(2x−1)⋅(x+13)n\sum_{n=0}^{\infty} (2x-1) \cdot \left(\frac{x+1}{3}\right)^n converge, and what is its sum when it converges?

  1. Converges for −4<x<2-4 < x < 2; sum is 3(2x−1)2−x\frac{3(2x-1)}{2-x} (correct answer)
  2. Converges for −4<x<2-4 < x < 2; sum is 6(2x−1)2−x\frac{6(2x-1)}{2-x}
  3. Converges for −2<x<4-2 < x < 4; sum is 3(2x−1)2−x\frac{3(2x-1)}{2-x}
  4. Converges for −4<x<2-4 < x < 2; sum is 9(2x−1)2−x\frac{9(2x-1)}{2-x}
Explanation: This is an infinite geometric series with first term a=2x−1a = 2x-1 and common ratio r=x+13r = \frac{x+1}{3}. For convergence, we need ∣r∣<1|r| < 1, which means ∣x+13∣<1\left|\frac{x+1}{3}\right| < 1. This gives ∣x+1∣<3|x+1| < 3, so −3<x+1<3-3 < x+1 < 3, which means −4<x<2-4 < x < 2. When the series converges, its sum is S=a1−r=2x−11−x+13=2x−13−(x+1)3=2x−12−x3=3(2x−1)2−xS = \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 3

An infinite geometric series has first term a=18a = 18 and sum S=24S = 24. What is the common ratio rr?

  1. r=14r = \frac{1}{4} (correct answer)
  2. r=13r = \frac{1}{3}
  3. r=34r = \frac{3}{4}
  4. r=23r = \frac{2}{3}
Explanation: For an infinite geometric series with ∣r∣<1|r| < 1, the sum is S=a1−rS = \frac{a}{1-r}. Substituting the given values: 24=181−r24 = \frac{18}{1-r}. Solving for rr: 24(1−r)=1824(1-r) = 18, so 24−24r=1824 - 24r = 18, which gives 24r=624r = 6, and therefore r=14r = \frac{1}{4}. We can verify: ∣14∣<1|\frac{1}{4}| < 1 (convergence condition satisfied) and S=181−14=1834=24S = \frac{18}{1-\frac{1}{4}} = \frac{18}{\frac{3}{4}} = 24 ✓. Choice B gives r=13r = \frac{1}{3}: S=181−13=1823=27≠24S = \frac{18}{1-\frac{1}{3}} = \frac{18}{\frac{2}{3}} = 27 \neq 24. Choice C gives r=34r = \frac{3}{4}: S=181−34=1814=72≠24S = \frac{18}{1-\frac{3}{4}} = \frac{18}{\frac{1}{4}} = 72 \neq 24. Choice D gives r=23r = \frac{2}{3}: S=181−23=1813=54≠24S = \frac{18}{1-\frac{2}{3}} = \frac{18}{\frac{1}{3}} = 54 \neq 24.

Question 4

Consider the series ∑k=273⋅2k−2\sum_{k=2}^{7} 3 \cdot 2^{k-2}. This can be rewritten as a geometric series. What is its sum?

  1. S=189S = 189 (correct answer)
  2. S=192S = 192
  3. S=195S = 195
  4. S=198S = 198
Explanation: Let's rewrite this series by substituting j=k−2j = k - 2, so when k=2k = 2, j=0j = 0 and when k=7k = 7, j=5j = 5. The series becomes ∑j=053⋅2j\sum_{j=0}^{5} 3 \cdot 2^j. This is a geometric series with first term a=3⋅20=3a = 3 \cdot 2^0 = 3, common ratio r=2r = 2, and 6 terms (from j=0j = 0 to j=5j = 5). Using Sn=a⋅rn−1r−1S_n = a \cdot \frac{r^n - 1}{r - 1}: S6=3⋅26−12−1=3⋅64−11=3⋅63=189S_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⋅20=33 \cdot 2^0 = 3, 3⋅21=63 \cdot 2^1 = 6, 3⋅22=123 \cdot 2^2 = 12, 3⋅23=243 \cdot 2^3 = 24, 3⋅24=483 \cdot 2^4 = 48, 3⋅25=963 \cdot 2^5 = 96. Sum: 3+6+12+24+48+96=1893 + 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.