College Algebra Quiz: Geometric Sequences Explicit And Recursive Forms
7 questions · exam conditions
0:00
Geometric Sequences Explicit And Recursive FormsQuestion 1 of 7

Consider a geometric sequence where a2=15a_2 = 15 and a5=−405a_5 = -405. What is the recursive relationship for this sequence?

an+1=3ana_{n+1} = 3a_n
an+1=−3ana_{n+1} = -3a_n
an+1=−13ana_{n+1} = -\frac{1}{3}a_n
an+1=13ana_{n+1} = \frac{1}{3}a_n
← Back to quizzes

College Algebra Quiz

College Algebra Quiz: Geometric Sequences Explicit And Recursive Forms

Practice Geometric Sequences Explicit And Recursive Forms 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 Geometric Sequences Explicit And Recursive Forms, 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.

All questions

Question 1

Consider a geometric sequence where a2=15a_2 = 15 and a5=−405a_5 = -405. What is the recursive relationship for this sequence?

  1. an+1=3ana_{n+1} = 3a_n
  2. an+1=−3ana_{n+1} = -3a_n (correct answer)
  3. an+1=−13ana_{n+1} = -\frac{1}{3}a_n
  4. an+1=13ana_{n+1} = \frac{1}{3}a_n
Explanation: When you encounter a geometric sequence problem with two given terms, your goal is to find the common ratio that connects consecutive terms through multiplication. In a geometric sequence, each term equals the previous term multiplied by the common ratio rr. Since a2=15a_2 = 15 and a5=−405a_5 = -405, you can use the relationship a5=a2⋅r3a_5 = a_2 \cdot r^3 (since you multiply by rr three times to go from the 2nd to 5th term). Substituting the known values: −405=15⋅r3-405 = 15 \cdot r^3 Solving for rr: r3=−40515=−27r^3 = \frac{-405}{15} = -27 Therefore: r=−273=−3r = \sqrt[3]{-27} = -3 The recursive relationship is an+1=r⋅an=−3ana_{n+1} = r \cdot a_n = -3a_n, which is answer B. Let's check why the other options are incorrect: A) an+1=3ana_{n+1} = 3a_n: This uses r=3r = 3, which would give a5=15⋅33=405a_5 = 15 \cdot 3^3 = 405, not −405-405. The sign is wrong. C) an+1=−13ana_{n+1} = -\frac{1}{3}a_n: This uses r=−13r = -\frac{1}{3}, making the sequence decrease in magnitude rather than increase. D) an+1=13ana_{n+1} = \frac{1}{3}a_n: This uses r=13r = \frac{1}{3}, which would make a5=59a_5 = \frac{5}{9}, far too small and positive. Study tip: Always verify your common ratio by checking that it produces the correct second given term. Calculate a2⋅r(n−2)a_2 \cdot r^{(n-2)} to find ana_n, and pay careful attention to negative signs—they often determine the correct answer in geometric sequence problems.

Question 2

A geometric sequence is defined recursively by a1=7a_1 = 7 and an+1=kana_{n+1} = ka_n where kk is a constant. If a5=112a_5 = 112, which explicit formula correctly represents this sequence?

  1. an=7⋅2na_n = 7 \cdot 2^n
  2. an=7⋅2n−1a_n = 7 \cdot 2^{n-1} (correct answer)
  3. an=14⋅2n−1a_n = 14 \cdot 2^{n-1}
  4. an=7⋅(2)n−1a_n = 7 \cdot (\sqrt{2})^{n-1}
Explanation: When you encounter a geometric sequence problem, you need to connect the recursive definition with the explicit formula. A geometric sequence has the form an=a1⋅rn−1a_n = a_1 \cdot r^{n-1} where a1a_1 is the first term and rr is the common ratio. First, let's find the common ratio kk. Since a1=7a_1 = 7 and a5=112a_5 = 112, we can use the relationship a5=a1⋅k4a_5 = a_1 \cdot k^4: 112=7⋅k4112 = 7 \cdot k^4 k4=16k^4 = 16 k=2k = 2 Now we can write the explicit formula: an=7⋅2n−1a_n = 7 \cdot 2^{n-1}. Let's verify: a5=7⋅25−1=7⋅24=7⋅16=112a_5 = 7 \cdot 2^{5-1} = 7 \cdot 2^4 = 7 \cdot 16 = 112 ✓ Looking at the wrong answers: Choice A gives an=7⋅2na_n = 7 \cdot 2^n, which would make a5=7⋅25=224a_5 = 7 \cdot 2^5 = 224—this formula has an extra factor of 2 because it uses 2n2^n instead of 2n−12^{n-1}. Choice C gives an=14⋅2n−1a_n = 14 \cdot 2^{n-1}, which would make a1=14a_1 = 14, contradicting the given condition that a1=7a_1 = 7. Choice D gives an=7⋅(2)n−1a_n = 7 \cdot (\sqrt{2})^{n-1}, which would make a5=7⋅(2)4=7⋅4=28a_5 = 7 \cdot (\sqrt{2})^4 = 7 \cdot 4 = 28—this uses the wrong common ratio. The answer is B. Study tip: Always remember that the standard explicit formula for geometric sequences uses rn−1r^{n-1}, not rnr^n, because when n=1n=1, you want r0=1r^0 = 1 to get back to a1a_1.

Question 3

A geometric sequence has the property that a3=24a_3 = 24 and a6=−648a_6 = -648. If this sequence is defined recursively as an+1=r⋅ana_{n+1} = r \cdot a_n, what is the value of the common ratio rr?

  1. −3-3 (correct answer)
  2. 33
  3. −13-\frac{1}{3}
  4. 13\frac{1}{3}
Explanation: In a geometric sequence, a6=a3⋅r6−3=a3⋅r3a_6 = a_3 \cdot r^{6-3} = a_3 \cdot r^3. Substituting the given values: −648=24⋅r3-648 = 24 \cdot r^3, so r3=−64824=−27r^3 = \frac{-648}{24} = -27. Taking the cube root: r=−273=−3r = \sqrt[3]{-27} = -3. We can verify: if r=−3r = -3 and a3=24a_3 = 24, then a4=24(−3)=−72a_4 = 24(-3) = -72, a5=−72(−3)=216a_5 = -72(-3) = 216, and a6=216(−3)=−648a_6 = 216(-3) = -648 ✓. Choice B gives r3=27r^3 = 27, which would make a6=648a_6 = 648 (positive). Choices C and D would give much smaller values for a6a_6.

Question 4

The recursive formula for a geometric sequence is given by a1=5a_1 = 5 and an+1=−2ana_{n+1} = -2a_n for n≥1n \geq 1. What is the explicit formula for the nnth term?

  1. an=5(−2)n−1a_n = 5(-2)^{n-1} (correct answer)
  2. an=5(−2)na_n = 5(-2)^n
  3. an=−10⋅2n−1a_n = -10 \cdot 2^{n-1}
  4. an=5⋅2n−1a_n = 5 \cdot 2^{n-1}
Explanation: From the recursive formula an+1=−2ana_{n+1} = -2a_n, we can identify that the common ratio is r=−2r = -2. The general form of a geometric sequence is an=a1⋅rn−1a_n = a_1 \cdot r^{n-1}. Substituting a1=5a_1 = 5 and r=−2r = -2, we get an=5(−2)n−1a_n = 5(-2)^{n-1}. We can verify: a1=5(−2)0=5a_1 = 5(-2)^0 = 5, a2=5(−2)1=−10a_2 = 5(-2)^1 = -10, a3=5(−2)2=20a_3 = 5(-2)^2 = 20. Choice B is incorrect because it uses (−2)n(-2)^n instead of (−2)n−1(-2)^{n-1}, giving a1=5(−2)1=−10a_1 = 5(-2)^1 = -10. Choice C incorrectly separates the negative sign. Choice D ignores the negative in the common ratio.

Question 5

A geometric sequence satisfies the recursive relation an+1=0.8ana_{n+1} = 0.8a_n with a4=40.96a_4 = 40.96. What is the explicit formula for this sequence?

  1. an=80(0.8)na_n = 80(0.8)^n
  2. an=64(0.8)n−1a_n = 64(0.8)^{n-1}
  3. an=80(0.8)n−1a_n = 80(0.8)^{n-1} (correct answer)
  4. an=51.2(0.8)n−1a_n = 51.2(0.8)^{n-1}
Explanation: When you encounter a geometric sequence with a recursive relation, you need to identify the common ratio and find the first term to write the explicit formula in the standard form an=a1⋅rn−1a_n = a_1 \cdot r^{n-1}. From the recursive relation an+1=0.8ana_{n+1} = 0.8a_n, you can see that the common ratio is r=0.8r = 0.8. To find the first term, work backwards from the given information a4=40.96a_4 = 40.96. Since each term is obtained by multiplying the previous term by 0.8, you can find earlier terms by dividing by 0.8: a3=a40.8=40.960.8=51.2a_3 = \frac{a_4}{0.8} = \frac{40.96}{0.8} = 51.2 a2=a30.8=51.20.8=64a_2 = \frac{a_3}{0.8} = \frac{51.2}{0.8} = 64 a1=a20.8=640.8=80a_1 = \frac{a_2}{0.8} = \frac{64}{0.8} = 80 Therefore, the explicit formula is an=80(0.8)n−1a_n = 80(0.8)^{n-1}, which is choice C. Choice A uses the correct first term (80) but has the exponent as nn instead of n−1n-1, which would make a1=80(0.8)1=64a_1 = 80(0.8)^1 = 64 instead of 80. Choice B incorrectly uses 64 as the coefficient, which would be a2a_2 rather than a1a_1. Choice D uses 51.2 as the coefficient, which represents a3a_3 instead of a1a_1. Remember: the standard form an=a1⋅rn−1a_n = a_1 \cdot r^{n-1} always uses the first term as the coefficient and n−1n-1 as the exponent. When given information about a later term, always work backwards to find a1a_1.

Question 6

A geometric sequence has explicit formula an=48⋅(14)n−1a_n = 48 \cdot (\frac{1}{4})^{n-1}. Starting from this explicit form, what is the equivalent recursive definition?

  1. a1=48a_1 = 48, an+1=4ana_{n+1} = 4a_n for n≥1n \geq 1
  2. a1=12a_1 = 12, an+1=14ana_{n+1} = \frac{1}{4}a_n for n≥1n \geq 1
  3. a1=48a_1 = 48, an+1=14ana_{n+1} = \frac{1}{4}a_n for n≥1n \geq 1 (correct answer)
  4. a1=192a_1 = 192, an+1=14ana_{n+1} = \frac{1}{4}a_n for n≥1n \geq 1
Explanation: When converting between explicit and recursive forms of geometric sequences, you need to identify two key components: the first term and the common ratio that connects consecutive terms. From the explicit formula an=48⋅(14)n−1a_n = 48 \cdot (\frac{1}{4})^{n-1}, you can find the first term by substituting n=1n = 1: a1=48⋅(14)1−1=48⋅(14)0=48⋅1=48a_1 = 48 \cdot (\frac{1}{4})^{1-1} = 48 \cdot (\frac{1}{4})^0 = 48 \cdot 1 = 48. The common ratio is the base of the exponential term, which is 14\frac{1}{4}. This means each term is 14\frac{1}{4} times the previous term, so an+1=14ana_{n+1} = \frac{1}{4}a_n. Choice C correctly identifies both components: a1=48a_1 = 48 and an+1=14ana_{n+1} = \frac{1}{4}a_n. Choice A has the correct first term but uses an+1=4ana_{n+1} = 4a_n, which would make the sequence increase rather than decrease. This represents the reciprocal of the actual common ratio. Choice B uses the wrong first term (a1=12a_1 = 12 instead of 48). This might come from incorrectly calculating 48⋅1448 \cdot \frac{1}{4}, but remember that when n=1n = 1, the exponent becomes zero, not one. Choice D has the correct recursive relationship but starts with a1=192a_1 = 192. This appears to come from multiplying 48 by 4, perhaps confusing the direction of the ratio. Study tip: Always verify your recursive formula by computing the first few terms both ways. The explicit formula an=a1⋅rn−1a_n = a_1 \cdot r^{n-1} directly shows you a1a_1 and rr, which become the building blocks of your recursive definition.

Question 7

Consider the geometric sequence defined by an=3⋅(12)n−1a_n = 3 \cdot (\frac{1}{2})^{n-1}. If we write this sequence recursively starting with a1=ka_1 = k, what is the correct recursive relationship?

  1. a1=32a_1 = \frac{3}{2}, an+1=2ana_{n+1} = 2a_n for n≥1n \geq 1
  2. a1=32a_1 = \frac{3}{2}, an+1=12ana_{n+1} = \frac{1}{2}a_n for n≥1n \geq 1
  3. a1=3a_1 = 3, an+1=2ana_{n+1} = 2a_n for n≥1n \geq 1
  4. a1=3a_1 = 3, an+1=12ana_{n+1} = \frac{1}{2}a_n for n≥1n \geq 1 (correct answer)
Explanation: When converting between explicit and recursive formulas for geometric sequences, you need to identify two key components: the first term and the common ratio that connects consecutive terms. Let's find the first term by substituting n=1n = 1 into the given formula: a1=3⋅(12)1−1=3⋅(12)0=3⋅1=3a_1 = 3 \cdot (\frac{1}{2})^{1-1} = 3 \cdot (\frac{1}{2})^0 = 3 \cdot 1 = 3. So our recursive formula must start with a1=3a_1 = 3. For the recursive relationship, we need the common ratio. In any geometric sequence, each term equals the previous term multiplied by the common ratio. Looking at our formula an=3⋅(12)n−1a_n = 3 \cdot (\frac{1}{2})^{n-1}, the base 12\frac{1}{2} is our common ratio. This means an+1=12⋅ana_{n+1} = \frac{1}{2} \cdot a_n. Let's verify: a1=3a_1 = 3, a2=3⋅12=32a_2 = 3 \cdot \frac{1}{2} = \frac{3}{2}, a3=32⋅12=34a_3 = \frac{3}{2} \cdot \frac{1}{2} = \frac{3}{4}. Using our explicit formula: a2=3⋅(12)1=32a_2 = 3 \cdot (\frac{1}{2})^1 = \frac{3}{2} ✓. Choice A uses the wrong first term (32\frac{3}{2} instead of 33) and wrong ratio (22 instead of 12\frac{1}{2}). Choice B has the correct ratio but wrong first term. Choice C has the correct first term but uses ratio 22, which would make the sequence grow instead of shrink. Only choice D correctly identifies both a1=3a_1 = 3 and an+1=12ana_{n+1} = \frac{1}{2}a_n. Study tip: Always check your recursive formula by computing the first few terms both ways—you should get identical sequences regardless of which formula you use.