Algebra 3 Quiz: Geometric Sequences And Series
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Geometric Sequences And SeriesQuestion 1 of 12

In a geometric sequence, a2=6a_2=6 and a5=48a_5=48. What is a7a_7?

192192
9696
384384
4848
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Algebra 3 Quiz

Algebra 3 Quiz: Geometric Sequences And Series

Practice Geometric Sequences And Series in Algebra 3 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 And Series, giving you a quick way to practice the rules, question types, and explanations that matter most for Algebra 3.

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

In a geometric sequence, a2=6a_2=6 and a5=48a_5=48. What is a7a_7?

  1. 192192 (correct answer)
  2. 9696
  3. 384384
  4. 4848
Explanation: Whenever you see a geometric sequence, isolate the common ratio rr first. Between a2=6a_2=6 and a5=48a_5=48, there are 3 steps, so 6r3=48⇒r3=8⇒r=2.6r^3=48 \quad\Rightarrow\quad r^3=8 \quad\Rightarrow\quad r=2. Now a7a_7 is 2 steps after a5a_5, so multiply by r2r^2: a7=48⋅22=48⋅4=192.a_7=48\cdot 2^2=48\cdot4=192. That is the correct value. The choice 9696 is one step too soon — it is actually a6=48⋅2a_6=48\cdot2, not a7a_7. The choice 384384 comes from multiplying by 23=82^3=8 instead of 22=42^2=4; that would land on a8a_8, or it happens if you count terms from a5a_5 to a7a_7 inclusively rather than counting the gaps between them. The choice 4848 simply repeats a5a_5, ignoring the common ratio entirely. Remember: in a geometric sequence, the exponent is the number of gaps, not the number of terms listed. To go from a5a_5 to a7a_7 is 2 gaps, so use r2r^2. Always confirm how many steps you are moving before multiplying.

Question 2

Find the sum of the first 8 terms of the geometric sequence 3,  −6,  12,  −24,  …3,\;-6,\;12,\;-24,\;\ldots.

  1. −255-255 (correct answer)
  2. 255255
  3. 129129
  4. −129-129
Explanation: Whenever you see a geometric sequence, first identify the common ratio by dividing any term by the previous one. Here −6/3=−2-6/3=-2, so each term is multiplied by −2-2. The sum of the first nn terms is Sn=a11−rn1−r.S_n=a_1\frac{1-r^n}{1-r}. With a1=3a_1=3, r=−2r=-2, and n=8n=8, S8=31−(−2)81−(−2)=31−2563=−255.S_8=3\frac{1-(-2)^8}{1-(-2)} =3\frac{1-256}{3} =-255. So the 8-term sum is −255-255. Notice (−2)8(-2)^8 is positive 256256, not −128-128; using the 7th power gives the sum of only 7 terms, which is 129129. That is the trap behind the 129129 choice: it stops one term too soon. The choice 255255 has the right magnitude but the wrong sign; because the 8th term is −384-384, adding it to the 7-term sum 129129 gives −255-255, not 255255. The choice −129-129 is likely the result of taking the 7-term sum 129129 and flipping its sign, or mixing up the formula's numerator and denominator. To stay safe, use the formula with n=8n=8, and check whether an even number of alternating terms should make the total negative.

Question 3

The repeating decimal 0.272727…0.272727\ldots can be written as an infinite geometric series. Which fraction equals this decimal?

  1. 27100\frac{27}{100}
  2. 311\frac{3}{11} (correct answer)
  3. 518\frac{5}{18}
  4. 14\frac{1}{4}
Explanation: Whenever you see a repeating decimal, think of it as an infinite geometric series whose common ratio is a power of 110\frac{1}{10}. Here, 0.272727…=0.27+0.0027+0.000027+⋯0.272727\ldots = 0.27 + 0.0027 + 0.000027 + \cdots This is a geometric series with first term a=27100a=\frac{27}{100} and common ratio r=1100r=\frac{1}{100}. Its sum is a1−r=271001−1100=2710099100=2799=311.\frac{a}{1-r}=\frac{\frac{27}{100}}{1-\frac{1}{100}} =\frac{\frac{27}{100}}{\frac{99}{100}} =\frac{27}{99} =\frac{3}{11}. So 311\frac{3}{11} is the correct fraction. The choice 27100\frac{27}{100} is just the first two decimal places, not the full repeating value — it equals 0.270.27, not 0.272727…0.272727\ldots. The choice 518\frac{5}{18} equals 0.2777…0.2777\ldots, which repeats a single 77, not the two-digit block 2727. And 14\frac{1}{4} equals 0.250.25, which is close but definitely not the repeating pattern. A quick study tip: for a decimal with a repeating block of length nn, write the block over nn nines and simplify. Since 2727 repeats, 2799=311\frac{27}{99}=\frac{3}{11}. This shortcut saves time and helps you check your geometric-series work.

Question 4

A ball is dropped from a height of 8 meters. On each bounce it rises to 60%60\% of the height from which it just fell. What total vertical distance does the ball travel before coming to rest?

  1. 12.812.8 m
  2. 2020 m
  3. 2424 m
  4. 3232 m (correct answer)
Explanation: Whenever you see a bouncing ball or any repeated-motion distance problem, think geometric series. The key is to track every segment: the ball drops, then rises and falls repeatedly until the bounces become negligible. The initial drop is 88 meters. After that, each bounce reaches 60%60\% of the previous height, so the rebound heights are 8(0.6),8(0.6)2,8(0.6)3,…8(0.6), 8(0.6)^2, 8(0.6)^3, \dots. For each rebound height, the ball travels up that distance and then down that same distance, so each contributes twice its height. The total distance is therefore 8+2(8(0.6)+8(0.6)2+8(0.6)3+⋯ ).8 + 2\left(8(0.6) + 8(0.6)^2 + 8(0.6)^3 + \cdots\right). The infinite sum in parentheses is a geometric series with first term 8(0.6)=4.88(0.6)=4.8 and ratio 0.60.6. Its sum is 4.81−0.6=12\frac{4.8}{1-0.6}=12. Twice that is 2424, and adding the initial 88 gives 8+24=32 m.8 + 24 = 32\text{ m}. So 3232 m is correct. The wrong answers show common traps. 12.812.8 m is just the first drop plus the first rise, ignoring all later bounces. 2020 m is the sum of only the falling distances 8+4.8+2.88+⋯8+4.8+2.88+\cdots, forgetting that every upward bounce also counts. 2424 m correctly sums all the up-and-down bounce distances after the first drop but forgets to include the initial 88-meter drop. Your takeaway: physically list out the motion first — down, up, down, up — then form a geometric series. Include the initial drop, and remember to double every rebound because each bounce has both an up and a down segment.

Question 5

If x−2,  2x+1,  4x+5x-2,\;2x+1,\;4x+5 are consecutive terms of a geometric sequence, what is the value of xx?

  1. −711-\frac{7}{11}
  2. −1-1
  3. −113-\frac{11}{3}
  4. −117-\frac{11}{7} (correct answer)
Explanation: When you see consecutive terms of a geometric sequence, remember that the ratio between each pair of consecutive terms is constant. So here, 2x+1x−2=4x+52x+1\frac{2x+1}{x-2} = \frac{4x+5}{2x+1}. Cross-multiplying gives (2x+1)2=(x−2)(4x+5)(2x+1)^2 = (x-2)(4x+5). Expand carefully: 4x2+4x+1=4x2−3x−104x^2+4x+1 = 4x^2-3x-10. Cancel the 4x24x^2 terms: 4x+1=−3x−104x+1 = -3x-10, so 7x=−117x = -11, and x=−117x = -\frac{11}{7}. This checks: the terms become −257,−157,−97-\frac{25}{7}, -\frac{15}{7}, -\frac{9}{7}, each multiplied by 35\frac35. The other choices come from common traps. −113-\frac{11}{3} is what you get if you expand (2x+1)2(2x+1)^2 incorrectly as 4x2+14x^2+1, dropping the middle term 4x4x. −1-1 makes the sequence arithmetic, not geometric: −3,−1,1-3, -1, 1 have a common difference, not a common ratio. −711-\frac{7}{11} is the reciprocal of the correct answer, suggesting the equation 7x=−117x=-11 was solved by flipping the fraction instead of dividing by 7. On exam day, set up the ratio equality first, then expand carefully—especially when squaring binomials. A quick check of the three terms with your answer confirms whether the common ratio is constant.

Question 6

The nnth partial sum of a geometric series is Sn=3(2n−1)S_n=3(2^n-1). What is the fourth term a4a_4?

  1. 4848
  2. 4545
  3. 2424 (correct answer)
  4. 2121
Explanation: When you are given a formula for the partial sum SnS_n, you are being tested on the relationship between a running total and a single term. The key move is to remember that the nnth term is the difference between consecutive partial sums: an=Sn−Sn−1a_n = S_n - S_{n-1} For this geometric series, S4=3(24−1)=3(16−1)=45S_4 = 3(2^4-1)=3(16-1)=45 and S3=3(23−1)=3(8−1)=21S_3 = 3(2^3-1)=3(8-1)=21 So the fourth term is a4=S4−S3=45−21=24.a_4 = S_4 - S_3 = 45 - 21 = 24. The choice 4545 is the trap of reporting the partial sum S4S_4 itself, not the fourth term. The choice 2121 is S3S_3, the sum of the first three terms, not the fourth term. The choice 4848 comes from incorrectly using a4=3(24)=48a_4 = 3(2^4)=48, treating the exponent as nn instead of n−1n-1 for a term; even in the sum formula, the exponent inside is nn, but the individual term is 3⋅233 \cdot 2^{3}. A quick study tip: whenever a question gives you partial sums, immediately write an=Sn−Sn−1a_n = S_n - S_{n-1}. That single formula converts a potentially tricky series question into a simple subtraction problem.

Question 7

An infinite geometric series has sum 1212 and common ratio −13-\frac13. What is its second term?

  1. −4-4
  2. 169\frac{16}{9}
  3. −163-\frac{16}{3} (correct answer)
  4. 1616
Explanation: Whenever an infinite geometric series gives you its sum and common ratio, your first move is to recover the first term using the formula S=a1−rS = \frac{a}{1-r}. Here S=12S=12 and r=−13r=-\frac13, so a=S(1−r)=12(1+13)=16.a = S(1-r) = 12\left(1+\frac13\right)=16. That 1616 is the first term, not the answer. The second term is the first term multiplied by the common ratio: ar=16⋅(−13)=−163.ar = 16 \cdot \left(-\frac13\right) = -\frac{16}{3}. Now see why the other choices miss the mark. −4-4 comes from multiplying the sum 1212 by the ratio −13-\frac13; that conflates the total sum with the first term. 169\frac{16}{9} is actually the third term, ar2=16⋅19ar^2 = 16 \cdot \frac19; it confuses the term position with the exponent and also drops the negative sign. 1616 is the first term you computed correctly, but the question asks for the second term, not the first. On exam day, write the sum formula first, solve for aa, then take one more step to arar. Remember: the infinite sum formula gives you the first term, but the question may ask for a different term. Always check the requested term number and use arn−1ar^{n-1}.

Question 8

Which expression gives the nnth term of the geometric sequence 12,  −1,  2,  −4,  …\frac12,\;-1,\;2,\;-4,\;\ldots?

  1. an=−12(−2)n−1a_n=-\frac12(-2)^{n-1}
  2. an=12(−2)n−1a_n=\frac12(-2)^{n-1} (correct answer)
  3. an=12(−2)na_n=\frac12(-2)^n
  4. an=(−2)n−1a_n=(-2)^{n-1}
Explanation: Whenever you see a geometric sequence, your first move is to identify the first term a1a_1 and the common ratio rr. The nnth term is always an=a1rn−1a_n = a_1 r^{n-1}. Here the first term is 12\frac12, and dividing consecutive terms gives r=−2r=-2, since −1÷12=−2-1 \div \frac12 = -2, 2÷−1=−22 \div -1 = -2, and −4÷2=−2-4 \div 2 = -2. So the expression must be an=12(−2)n−1a_n=\frac12(-2)^{n-1}. The choice an=12(−2)n−1a_n=\frac12(-2)^{n-1} is correct because when n=1n=1, it gives 12(−2)0=12\frac12(-2)^0=\frac12, and then each next term multiplies by −2-2, producing −1,2,−4-1, 2, -4, and so on. Now look at the distractors. The expression an=−12(−2)n−1a_n=-\frac12(-2)^{n-1} has the wrong sign on the first term: at n=1n=1 it gives −12-\frac12, which is not the first term. The expression an=12(−2)na_n=\frac12(-2)^n uses an exponent of nn instead of n−1n-1; at n=1n=1 it gives 12(−2)=−1\frac12(-2)=-1, which is actually the second term, so it is shifted one position too late. Finally, an=(−2)n−1a_n=(-2)^{n-1} leaves out the initial factor 12\frac12; at n=1n=1 it gives 11, not 12\frac12. A quick study tip: always check your formula by plugging in n=1n=1. The exponent should be n−1n-1 so the first term comes out exactly as given, and don't forget the original starting value.

Question 9

How many terms are in the geometric sequence 3,  6,  12,  …,  7683,\;6,\;12,\;\ldots,\;768?

  1. 88
  2. 99 (correct answer)
  3. 1010
  4. 77
Explanation: Whenever you see a geometric sequence, start with the general term formula: an=a1rn−1a_n = a_1 r^{n-1}. Here a1=3a_1 = 3, and each term doubles, so r=2r=2. You need the term number nn such that an=768a_n = 768. Set up the equation: 3(2n−1)=7683(2^{n-1}) = 768 Divide by 3: 2n−1=2562^{n-1} = 256 Since 256=28256 = 2^8, you get n−1=8n-1 = 8, so n=9n = 9. The sequence is 3,6,12,24,48,96,192,384,7683, 6, 12, 24, 48, 96, 192, 384, 768, which has exactly 9 terms. The choice 88 is the exponent n−1n-1 — it counts the number of doublings after the first term, not the number of terms. The choice 77 likely comes from counting only the terms strictly between 33 and 768768, leaving out both endpoints. The choice 1010 comes from adding an extra term, such as continuing to 15361536 or counting the first term twice. The key habit: in a geometric sequence, the exponent on rr is always one less than the term number. So when you solve and get an exponent of 8, remember to add 1 back to find 9 terms. That off-by-one trap is exactly what this question tests.

Question 10

What is the sum of the first five terms of the geometric sequence 1,  −23,  49,  −827,  …1,\;-\frac{2}{3},\;\frac{4}{9},\;-\frac{8}{27},\;\ldots?

  1. 35\frac{3}{5}
  2. 1327\frac{13}{27}
  3. 5581\frac{55}{81} (correct answer)
  4. 21181\frac{211}{81}
Explanation: Whenever you see a geometric sequence, start by finding the common ratio rr by dividing any term by the one before it. Here, r=−2/31=−23.r=\frac{-2/3}{1}=-\frac{2}{3}. The sequence alternates signs, and the first five terms are 1,  −23,  49,  −827,  16811,\;-\frac23,\;\frac49,\;-\frac8{27},\;\frac{16}{81}. You can either add them directly using denominator 8181, 81−54+36−24+1681=5581,\frac{81-54+36-24+16}{81}=\frac{55}{81}, or use the finite geometric sum formula S5=1(1−(−2/3)5)1−(−2/3)=1+3224353=275243⋅35=5581.S_5=\frac{1\left(1-(-2/3)^5\right)}{1-(-2/3)} =\frac{1+\frac{32}{243}}{\frac53} =\frac{275}{243}\cdot\frac35 =\frac{55}{81}. So the sum is 5581\frac{55}{81}. As for the other answers, 35\frac35 is the sum of the infinite geometric series, since 1/(1+23)=351/(1+\frac23)=\frac35; but the question asks for the first five terms only. 1327\frac{13}{27} is the sum of only the first four terms, so it is the result of stopping one term too early. 21181\frac{211}{81} comes from adding the terms as if all signs were positive, ignoring the fact that the common ratio is negative. In short: always include the sign of rr, and make sure you use the exact number of terms requested.

Question 11

The second term of an infinite geometric series is −15-15 and the third term is 7.57.5. What is the sum of the series?

  1. 2020 (correct answer)
  2. −60-60
  3. −10-10
  4. 55
Explanation: Whenever you see an infinite geometric series, the key is to find the common ratio first, then the first term. Here, the third term divided by the second term gives the ratio: r=7.5−15=−0.5r=\frac{7.5}{-15}=-0.5 Since the second term is ar=−15ar=-15, we have a(−0.5)=−15⇒a=30a(-0.5)=-15 \quad\Rightarrow\quad a=30 Now apply the infinite sum formula: S=a1−r=301−(−0.5)=301.5=20S=\frac{a}{1-r}=\frac{30}{1-(-0.5)}=\frac{30}{1.5}=20 So the sum is indeed 2020. Why the others miss the mark:
  • −60-60 comes from using r=0.5r=0.5 instead of −0.5-0.5, which ignores the alternating sign. That would give a=−30a=-30 and sum −60-60.
  • −10-10 treats the second term −15-15 as if it were the first term: −15/(1−(−0.5))=−10-15/(1-(-0.5))=-10. But you must divide by the ratio to recover the actual first term.
  • 55 treats the third term 7.57.5 as the first term: 7.5/(1−(−0.5))=57.5/(1-(-0.5))=5. This skips the step of finding aa entirely.
A good habit: always compute rr first, then use a given term to find aa before applying S=a1−rS=\frac{a}{1-r}. Also watch the sign of rr — a negative ratio means the series alternates, and ignoring that sign is a common trap.

Question 12

For which values of xx does the infinite series ∑n=1∞(3x−1)n−1\sum_{n=1}^{\infty}\left(\frac{3}{x-1}\right)^{n-1} converge, and what is its sum?

  1. −4<x<2-4<x<2; x−1x−4\frac{x-1}{x-4}
  2. x<−2x<-2 or x>4x>4; x−1x+4\frac{x-1}{x+4}
  3. −2<x<4-2<x<4; x−1x−4\frac{x-1}{x-4}
  4. x<−2x<-2 or x>4x>4; x−1x−4\frac{x-1}{x-4} (correct answer)
Explanation: Whenever you see an infinite series whose terms are powers of a fixed expression, think geometric series: it converges exactly when ∣r∣<1|r|<1, and its sum is a1−r\frac{a}{1-r}. Here the first term is (3x−1)0=1\left(\frac{3}{x-1}\right)^0=1, and the common ratio is r=3x−1r=\frac{3}{x-1}. So convergence requires ∣3x−1∣<1⇒∣x−1∣>3.\left|\frac{3}{x-1}\right|<1 \quad \Rightarrow \quad |x-1|>3. This means x−1>3x-1>3 or x−1<−3x-1<-3, giving the interval x<−2x<-2 or x>4x>4. For those xx, the sum is 11−3x−1=x−1x−4.\frac{1}{1-\frac{3}{x-1}}=\frac{x-1}{x-4}. So the correct choice is the one with interval x<−2x<-2 or x>4x>4 and sum x−1x−4\frac{x-1}{x-4}. The choice with interval −4<x<2-4<x<2 misplaces the center of the inequality: the condition is about distance from 11, not from −1-1, and it must be greater than 33. The choice with sum x−1x+4\frac{x-1}{x+4} uses the correct interval but adds the ratio in the denominator instead of subtracting it: 1−r1-r, not 1+r1+r. The choice −2<x<4-2<x<4 reverses the inequality; that interval actually makes ∣r∣>1|r|>1, so the series diverges there. Study tip: for any geometric series, write down aa and rr first. Then solve ∣r∣<1|r|<1 carefully, and simplify a1−r\frac{a}{1-r} separately. That keeps the interval and the sum from getting tangled together.