Algebra 2 Quiz: Deriving Applying The Geometric Series Formula
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Deriving Applying The Geometric Series FormulaQuestion 1 of 20

Compute the sum using the finite geometric series formula: 2+2(1.1)+2(1.1)2+⋯+2(1.1)9.2+2(1.1)+2(1.1)^2+\cdots+2(1.1)^9.

2(1−(1.1)9)1−1.1\dfrac{2\big(1-(1.1)^9\big)}{1-1.1}
2(1−(1.1)10)1−1.1\dfrac{2\big(1-(1.1)^{10}\big)}{1-1.1}
2(1−(1.1)10)1+1.1\dfrac{2\big(1-(1.1)^{10}\big)}{1+1.1}
2(1.1)92(1.1)^9
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Algebra 2 Quiz

Algebra 2 Quiz: Deriving Applying The Geometric Series Formula

Practice Deriving Applying The Geometric Series Formula in Algebra 2 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 Deriving Applying The Geometric Series Formula, giving you a quick way to practice the rules, question types, and explanations that matter most for Algebra 2.

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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

Compute the sum using the finite geometric series formula: 2+2(1.1)+2(1.1)2+⋯+2(1.1)9.2+2(1.1)+2(1.1)^2+\cdots+2(1.1)^9.

  1. 2(1−(1.1)9)1−1.1\dfrac{2\big(1-(1.1)^9\big)}{1-1.1}
  2. 2(1−(1.1)10)1−1.1\dfrac{2\big(1-(1.1)^{10}\big)}{1-1.1} (correct answer)
  3. 2(1−(1.1)10)1+1.1\dfrac{2\big(1-(1.1)^{10}\big)}{1+1.1}
  4. 2(1.1)92(1.1)^9
Explanation: This question tests your understanding of geometric series—the sum of terms from a geometric sequence—and how to derive and use the formula Sn=a1−rn1−rS_n = a \frac{1 - r^n}{1 - r} to calculate these sums efficiently. A geometric series is the sum of terms from a geometric sequence: here we have 2, 2(1.1), 2(1.1) squared, ..., 2(1.1) to the 9, where each term is multiplied by 1.1 to get the next. Rather than computing each power of 1.1 and adding, we use the formula Sn=a1−rn1−rS_n = a \frac{1 - r^n}{1 - r}. To identify parameters: a = 2 (first term), r = 1.1 (common ratio), and n = 10 (count carefully: from 2(1.1) to the 0 up to 2(1.1) to the 9 gives us 10 terms total—don't forget that (1.1) to the 0 = 1!). The formula gives S10=21−(1.1)101−1.1S_{10} = 2 \frac{1 - (1.1)^{10}}{1 - 1.1}. Choice B correctly shows this expression with n = 10. Choice A incorrectly uses n = 9, missing that we start with 2 = 2(1.1) to the 0 and end with 2(1.1) to the 9, giving 10 terms total. When the last term has exponent 9, there are 10 terms because we include the 0th power! The three-step geometric series strategy: (1) Count terms by looking at exponents: from r0r^0 to r9r^9 is 10 terms. (2) Identify a as the coefficient (here 2). (3) Apply the formula carefully. Remember: when terms go from r0r^0 to rkr^k, you have (k+1) terms total!

Question 2

For the geometric series 5+15+45+135+405,5+15+45+135+405, identify aa, rr, and nn, then find SnS_n.

  1. a=5, r=3, n=5, S5=605a=5,\ r=3,\ n=5,\ S_5=605 (correct answer)
  2. a=5, r=3, n=5, S5=600a=5,\ r=3,\ n=5,\ S_5=600
  3. a=15, r=3, n=5, S5=605a=15,\ r=3,\ n=5,\ S_5=605
  4. a=5, r=5, n=3, S3=125a=5,\ r=5,\ n=3,\ S_3=125
Explanation: This question tests your understanding of geometric series—the sum of terms from a geometric sequence—and identifying parameters to apply S_n = a(1 - rnr^n)/(1 - r). A geometric series is the sum of terms from a geometric sequence: here 5 + 15 + 45 + 135 + 405 (multiply by 3 each time, 5 terms). Rather than adding manually, use the formula with a=5, r=3, n=5: S_5 = 5(1 - 353^5)/(1 - 3) = 5(1 - 243)/(-2) = 5(-242)/(-2) = 605. Verify: 5+15=20, +45=65, +135=200, +405=605—perfect! Choice A correctly identifies a=5, r=3, n=5 and computes S_5=605. Choice C mistakes a=15 (second term) instead of first term 5; always start with the initial term! The three-step strategy: (1) List terms to find a (first), r (ratio), n (count), (2) Check r ≠1, (3) Substitute carefully. Mastering parameter ID builds confidence for any series—keep going!

Question 3

A ball is dropped from 1010 ft and each bounce reaches 80%80\% of the previous height. What is the total vertical distance traveled during the first 5 bounces (up and down), not including the initial drop?

Distance for bounce kk is 2(10)(0.8k)2(10)(0.8^k) for k=1,2,3,4,5k=1,2,3,4,5.

  1. 20(1−0.851−0.8)20\left(\dfrac{1-0.8^5}{1-0.8}\right)
  2. 16(1−0.851−0.8)16\left(\dfrac{1-0.8^5}{1-0.8}\right) (correct answer)
  3. 20(1−0.861−0.8)20\left(\dfrac{1-0.8^6}{1-0.8}\right)
  4. 20(1−0.851+0.8)20\left(\dfrac{1-0.8^5}{1+0.8}\right)
Explanation: This question tests your understanding of geometric series—the sum of terms from a geometric sequence—and adapting the formula S_n = a(1 - rnr^n) / (1 - r) for applications like bouncing ball distances. The total distance for 5 bounces is sum_{k=1}^5 210(0.8)^k = 20 sum_{k=1}^5 (0.8)^k, where sum = 0.8(1 - 0.850.8^5)/(1 - 0.8) since it starts from k=1, not 0. Rather than calculating each bounce, the formula handles the pattern: adjust to a=20*0.8=16, then S=16(1−0.851-0.8^5)/(1-0.8). Cancellation in derivation makes it reliable for decaying ratios like 0.8<1. Choice B correctly sets a=16, r=0.8, n=5 for the adjusted series. Choice A uses 20 as a but includes an extra term like sum from k=0, overestimating—remember to shift for starting at k=1! Factor out the first multiplier, confirm n by listing a few terms, and note |r|<1 means terms shrink—you're mastering these practical uses, great job!

Question 4

Derive the finite geometric series sum formula. Start with Sn=a+ar+ar2+⋯+arn−1.S_n=a+ar+ar^2+\cdots+ar^{n-1}. Multiply both sides by rr and subtract to show the telescoping cancellation. Which expression correctly results from subtracting rSnrS_n from SnS_n?​

  1. Sn−rSn=a−arnS_n-rS_n=a-ar^n (correct answer)
  2. Sn−rSn=a−arn−1S_n-rS_n=a-ar^{n-1}
  3. Sn−rSn=ar−arnS_n-rS_n=ar-ar^{n}
  4. Sn−rSn=a+arnS_n-rS_n=a+ar^n
Explanation: This question tests your understanding of geometric series—the sum of terms from a geometric sequence—and how to derive and use the formula S_n = a(1 - rnr^n)/(1 - r) to calculate these sums efficiently. The derivation uses a clever algebraic trick: start with S_n = a + ar + ar^2 + ... + ar^(n-1), then multiply both sides by r to get rS_n = ar + ar^2 + ar^3 + ... + ar^n. Now subtract the second equation from the first: S_n - rS_n = (a + ar + ar^2 + ... + ar^(n-1)) - (ar + ar2ar^2 + ar3ar^3 + ... + arnar^n). The magic happens with telescoping cancellation: ar cancels with ar, ar^2 cancels with ar^2, and so on, leaving only S_n - rS_n = a - ar^n. Choice A correctly shows this result after the cancellation. Choices B and C have incorrect terms remaining after cancellation—B suggests ar^(n-1) survives (but it cancels with ar^(n-1) from rS_n), while C incorrectly keeps ar instead of a; choice D has the wrong sign on ar^n. The key insight is that all middle terms cancel perfectly, leaving only the first term from S_n (which is a) and the negative of the last term from rS_n (which is −arn-ar^n). This telescoping cancellation is what makes the formula derivation work!

Question 5

A person deposits 100100 at the end of each month into an account that earns 1%1\% interest per month. After 6 deposits, the account balance (right after the 6th deposit) is 100+100(1.01)+100(1.01)2+⋯+100(1.01)5.100 + 100(1.01) + 100(1.01)^2 + \cdots + 100(1.01)^5. Which is the correct exact expression for this sum using the geometric series formula?

  1. 100 1−(1.01)61−1.01100\,\dfrac{1-(1.01)^6}{1-1.01} (correct answer)
  2. 100 1−(1.01)51−1.01100\,\dfrac{1-(1.01)^5}{1-1.01}
  3. 100 1−(1.01)61+1.01100\,\dfrac{1-(1.01)^6}{1+1.01}
  4. 100 1−(1.01)71−1.01100\,\dfrac{1-(1.01)^7}{1-1.01}
Explanation: This question tests your understanding of geometric series—the sum of terms from a geometric sequence—and how to derive and use the formula Sn=a1−rn1−rS_n = a \frac{1 - r^n}{1 - r} to calculate these sums efficiently. A geometric series models compound growth: here, each deposit earns 1% interest monthly, so earlier deposits grow more. The series 100+100(1.01)+100(1.01)2+⋯+100(1.01)5100 + 100(1.01) + 100(1.01)^2 + \cdots + 100(1.01)^5 represents the value of all deposits at the time of the 6th deposit. To identify parameters: a=100a = 100 (first term, the 6th deposit made just now), r=1.01r = 1.01 (each earlier deposit has grown by factor 1.01 per month), and n=6n = 6 (we have 6 deposits total). The most recent deposit hasn't earned interest yet (power 0), while the first deposit has compounded for 5 months (power 5). Applying the formula: S6=1001−(1.01)61−1.01S_6 = 100 \frac{1 - (1.01)^6}{1 - 1.01}. Choice A correctly shows this expression. Choice B incorrectly uses n=5n = 5—remember we're summing 6 deposits, even though the highest power is 5. The pattern is: 6th deposit (just made) = 100(1.01)0100(1.01)^0, 5th deposit = 100(1.01)1100(1.01)^1, ..., 1st deposit = 100(1.01)5100(1.01)^5. The three-step geometric series strategy for finance problems: (1) Identify which payment is "now" (usually has power 0). (2) Count all payments (here 6 deposits). (3) Apply formula with correct nn. Financial applications often have this "off by one" trap where nn terms means highest power is (n−1n-1)!

Question 6

A finite geometric series has first term a=12a=12, common ratio r=13r=\tfrac{1}{3}, and n=7n=7 terms. What is S7S_7? (Note: If ∣r∣<1|r|<1, the infinite sum would converge, but here you must find the finite sum.)

  1. 12(1−(13)7)1−13\dfrac{12\left(1-\left(\tfrac13\right)^7\right)}{1-\tfrac13} (correct answer)
  2. 12(1−(13)6)1−13\dfrac{12\left(1-\left(\tfrac13\right)^6\right)}{1-\tfrac13}
  3. 12(1−(13)7)1+13\dfrac{12\left(1-\left(\tfrac13\right)^7\right)}{1+\tfrac13}
  4. 12⋅712\cdot 7
Explanation: This question tests your understanding of geometric series—the sum of terms from a geometric sequence—and using the finite sum formula even when |r|<1 suggests infinite convergence. A geometric series is the sum of terms from a geometric sequence: here with a=12, r=1/3, n=7, so terms 12, 4, 4/3, etc., summing finitely. Rather than adding, use S_7 = 12(1 - (1/3)^7)/(1 - 1/3); note it's finite, not infinite. For |r|<1, infinite sum is a/(1-r), but here n=7 limits it. Choice A correctly applies the formula with n=7 and r^7. Choice B uses r^6 for n=6, perhaps miscounting terms—verify n=7! The three-step strategy: (1) Distinguish finite vs. infinite, identify parameters, (2) Use finite formula for given n, (3) Substitute accurately. Finite sums are key before infinity—wonderful job!

Question 7

A phone trade-in promotion gives you $400 today, but if you wait, the offer decreases by 15%15\% each week. If you wait 6 weeks, the total value of receiving the offer each week (hypothetically adding all weekly offers) is modeled by the finite geometric series 400+400(0.85)+400(0.85)2+⋯+400(0.85)5.400 + 400(0.85) + 400(0.85)^2 + \cdots + 400(0.85)^5. What is the exact sum in geometric-series form?

  1. 400⋅1−0.8551−0.85400\cdot \dfrac{1-0.85^5}{1-0.85}
  2. 400⋅(1−0.856)400\cdot (1-0.85^6)
  3. 400⋅1−0.8561+0.85400\cdot \dfrac{1-0.85^6}{1+0.85}
  4. 400⋅1−0.8561−0.85400\cdot \dfrac{1-0.85^6}{1-0.85} (correct answer)
Explanation: This question tests your understanding of geometric series—the sum of terms from a geometric sequence—and using Sn=a1−rn1−rS_n = a \frac{1 - r^n}{1 - r} to model decreasing values like trade-in offers. The series 400 + 400(0.85) + ... + 400(0.85)^5 has a=400, r=0.85, n=6 terms. Rather than adding, plug in: 4001−0.8561−0.85400 \frac{1 - 0.85^6}{1 - 0.85}, capturing the decay. Cancellation in derivation makes it ideal for ∣r∣<1|r|<1. Choice A correctly sets exponent 6 for 6 terms and denominator 1-0.85. Choice B uses 5, maybe forgetting the first term—count from 0 to 5 exponents. List terms to confirm n, compute carefully with decimals, and apply to real scenarios like this—you're developing versatile skills, keep it up!

Question 8

A ball is dropped from 1010 ft and bounces to 80%80\% of its previous height each time. What is the total vertical distance traveled during the first 55 bounces (up-and-down motion), not including the initial drop? (So include: up to first bounce height, down, up to second bounce height, down, ..., through the 5th bounce.)

  1. 101−0.8\dfrac{10}{1-0.8}
  2. 20(0.8+0.82+0.83+0.84+0.85)20\left(0.8+0.8^2+0.8^3+0.8^4+0.8^5\right) (correct answer)
  3. 10(1+0.8+0.82+0.83+0.84)10\left(1+0.8+0.8^2+0.8^3+0.8^4\right)
  4. 20(1+0.8+0.82+0.83+0.84)20\left(1+0.8+0.8^2+0.8^3+0.8^4\right)
Explanation: This question tests your understanding of geometric series—the sum of terms from a geometric sequence—and modeling physical scenarios like bounces with the formula. A geometric series is the sum of terms from a geometric sequence: here, each bounce height is 80% of previous, so distances are 28, 26.4, etc. (but exclude initial drop). Rather than listing, recognize total = 20 (0.8 + 0.820.8^2 + ... + 0.850.8^5), since 2100.8 for first, etc. The series has a=0.8, r=0.8, n=5 for the sum inside, multiplied by 20. Choice A correctly expresses the sum for 5 bounces, capturing the up-and-down pairs. Choice C includes an extra 1 in the sum, perhaps mistakenly adding initial drop— but the question excludes it! The three-step strategy: (1) Model the scenario to find the series (here, factor out the 20), (2) Identify inner a, r, n, (3) Write the expression. Applying geometrics to physics is awesome—you're doing great!

Question 9

A geometric series is the sum of terms of a geometric sequence. For the finite geometric series Sn=a+ar+ar2+⋯+arn−1,S_n=a+ar+ar^2+\cdots+ar^{n-1}, derive a formula for SnS_n (assume r≠1r\ne 1) by multiplying by rr and subtracting so that the middle terms cancel.

  1. Sn=a(1−rn−1)1−rS_n=\dfrac{a(1-r^{n-1})}{1-r}
  2. Sn=a(1−rn)1−rS_n=\dfrac{a(1-r^n)}{1-r} (correct answer)
  3. Sn=a(1−rn)1+rS_n=\dfrac{a(1-r^n)}{1+r}
  4. Sn=n(a+r)2S_n=\dfrac{n(a+r)}{2}
Explanation: This question tests your understanding of geometric series—the sum of terms from a geometric sequence—and how to derive the formula S_n = a(1 - rnr^n)/(1 - r) to calculate these sums efficiently. A geometric series is the sum of terms from a geometric sequence: if the sequence is a, ar, ar², ar³, the series is a + ar + ar² + ar³ (adding those terms). To derive the formula: (1) Write S_n = a + ar + ar² + ... + ar^(n-1), (2) Multiply both sides by r: rS_n = ar + ar² + ar³ + ... + ar^n, (3) Subtract second from first: S_n - rS_n = a - ar^n (all middle terms cancel!), (4) Factor: S_n(1 - r) = a(1 - rnr^n), (5) Divide by (1 - r): S_n = a(1 - rnr^n)/(1 - r). Choice B correctly shows this final formula with the proper exponent n (not n-1) on r in the numerator. Choice A incorrectly has r^(n-1) instead of r^n—remember when we multiply by r, the last term becomes ar^n, not ar^(n-1). The three-step strategy: identify that you're summing n terms (from ar^0 to ar^(n-1)), use the subtraction trick to cancel middle terms, and solve for S_n. This formula turns tedious addition into a simple calculation!

Question 10

A savings plan deposits money at the end of each month. The first deposit is $100, and each month the deposit is multiplied by 1.051.05 (a 5% increase). What is the total amount deposited after 66 months (ignore interest on the account itself)?​

  1. 100⋅1−(1.05)61−1.05100\cdot\dfrac{1-(1.05)^6}{1-1.05} (correct answer)
  2. 100⋅1−(1.05)51−1.05100\cdot\dfrac{1-(1.05)^5}{1-1.05}
  3. 100⋅1−(1.05)61.05100\cdot\dfrac{1-(1.05)^6}{1.05}
  4. 100(1.05)6100(1.05)^6
Explanation: This question tests your understanding of geometric series—the sum of terms from a geometric sequence—and how to derive and use the formula S_n = a(1 - rnr^n)/(1 - r) to calculate these sums efficiently. A geometric series forms from the deposits: first month $100, second month $100(1.05), third month $100(1.05)^2, and so on through the sixth month $100(1.05)^5. This gives us 6 terms total with a = 100 (first deposit), r = 1.05 (each deposit is 1.05 times the previous), and n = 6 (six monthly deposits). Applying the formula: S_6 = 100(1 - (1.05)^6)/(1 - 1.05), which matches choice A exactly. Choice B incorrectly uses n = 5 (only counting 5 deposits instead of 6); choice C has the wrong denominator (should be 1 - r = -0.05, not r = 1.05); choice D gives just the 7th deposit amount, not the sum of the first 6. The key insight: when counting deposits over 6 months, that's 6 terms in the series, and the last deposit is 100(1.05)^5 because we start with 100(1.05)^0 = 100. Real-world applications like savings plans often form geometric series when there's a constant percentage increase!

Question 11

Derive the finite geometric series sum formula for r≠1r \ne 1. Start with Sn=a+ar+ar2+⋯+arn−1.S_n = a + ar + ar^2 + \cdots + ar^{n-1}. Multiply both sides by rr and subtract to show the telescoping cancellation, then solve for SnS_n. Which expression is correct?

  1. Sn=a(1−rn)1+rS_n = \dfrac{a(1 - r^n)}{1 + r}
  2. Sn=a+(n−1)r2S_n = \dfrac{a + (n-1)r}{2}
  3. Sn=a(1−rn)1−rS_n = \dfrac{a(1 - r^n)}{1 - r} (correct answer)
  4. Sn=a(1−rn−1)1−rS_n = \dfrac{a(1 - r^{n-1})}{1 - r}
Explanation: This question tests your understanding of geometric series—the sum of terms from a geometric sequence—and how to derive the formula Sn=a(1−rn)1−rS_n = \dfrac{a(1 - r^n)}{1 - r} using telescoping cancellation. A geometric series is the sum of terms from a geometric sequence: if the sequence is a,ar,ar2,…,arn−1a, ar, ar^2, \dots, ar^{n-1}, the series is Sn=a+ar+ar2+⋯+arn−1S_n = a + ar + ar^2 + \dots + ar^{n-1}. Rather than adding manually, we derive the formula by a clever trick: multiply by rr and subtract to make terms cancel! To derive Sn=a(1−rn)1−rS_n = \dfrac{a(1 - r^n)}{1 - r}: (1) Write Sn=a+ar+ar2+⋯+arn−1S_n = a + ar + ar^2 + \dots + ar^{n-1}, (2) Multiply by rr: rSn=ar+ar2+⋯+arnr S_n = ar + ar^2 + \dots + ar^n, (3) Subtract: Sn−rSn=a−arnS_n - r S_n = a - ar^n, (4) Factor: Sn(1−r)=a(1−rn)S_n (1 - r) = a (1 - r^n), (5) Solve: Sn=a(1−rn)1−rS_n = \dfrac{a (1 - r^n)}{1 - r}; this telescoping is the key! Choice B correctly derives the formula through the subtraction and cancellation method, with the exponent nn in rnr^n. Choice A makes an error in the derivation by using rn−1r^{n-1} instead of rnr^n: when subtracting, the last term is −arn-ar^n, so 1−rn1 - r^n, not 1−rn−11 - r^{n-1}! The three-step derivation strategy: (1) Write SnS_n and rSnr S_n, (2) Subtract to cancel, (3) Solve for SnS_n. Why this is powerful: it handles any nn without listing terms—math magic at work!

Question 12

A manufacturing company plans to increase production by 8% each quarter for the next 5 quarters, starting with an initial production of 2,400 units in Quarter 1. What is the total number of units produced over all 5 quarters?

  1. 14,11214,112 units (correct answer)
  2. 15,36915,369 units
  3. 13,05613,056 units
  4. 14,59814,598 units
Explanation: This is a finite geometric series with first term a=2400a = 2400, common ratio r=1.08r = 1.08, and n=5n = 5 terms. Using the formula Sn=a⋅1−rn1−rS_n = a \cdot \frac{1-r^n}{1-r}: S5=2400⋅1−(1.08)51−1.08=2400⋅1−1.4693−0.08=2400⋅−0.4693−0.08=2400⋅5.8663=14,112S_5 = 2400 \cdot \frac{1-(1.08)^5}{1-1.08} = 2400 \cdot \frac{1-1.4693}{-0.08} = 2400 \cdot \frac{-0.4693}{-0.08} = 2400 \cdot 5.8663 = 14,112 units. Choice B uses r=0.08r = 0.08 instead of 1.081.08. Choice C calculates only the growth amount, not total production. Choice D incorrectly uses simple interest formula.

Question 13

A retirement account requires monthly payments. If the first payment is $800 and each subsequent payment increases by 2% from the previous payment, what is the total amount paid after 12 payments?

  1. $10,152.72
  2. $11,204.16
  3. $9,876.45
  4. $10,731.84 (correct answer)
Explanation: When you see a problem involving payments that increase by a fixed percentage each time, you're dealing with a geometric series. The key is recognizing that each payment is the previous payment multiplied by the same factor. Here, the first payment is $800, and each subsequent payment increases by 2%, meaning each payment equals the previous payment times 1.02. So the payments form the sequence: $800, $800(1.02), $800(1.02)², and so on. To find the total after 12 payments, you need the sum of a geometric series with first term $a=800a = 800 ,commonratio, common ratio r=1.02r = 1.02 ,and, and n=12n = 12 terms.Theformulais:terms. The formula is: Sn=a⋅1−rn1−rS_n = a \cdot \frac{1-r^n}{1-r} $ Substituting: S_{12} = 800 \cdot \frac{1-(1.02)^{12}}{1-1.02} = 800 \cdot \frac{1-1.2682}{-0.02} = 800 \cdot \frac{-0.2682}{-0.02} = 800 \cdot 13.4148 = $10,731.84 Choice A (10,152.72)likelyresultsfromusingsimpleinterestinsteadofcompoundgrowth,calculatingasifeachpaymentincreasedbyafixeddollaramountratherthanapercentage.ChoiceB(10,152.72) likely results from using simple interest instead of compound growth, calculating as if each payment increased by a fixed dollar amount rather than a percentage. Choice B (11,204.16) appears to use an incorrect formula or miscalculated the geometric series sum. Choice C ($9,876.45) is too low and might result from errors in the ratio calculation or using the wrong number of terms. Remember: percentage increases create geometric sequences, not arithmetic ones. Always identify whether the growth is by a fixed amount (arithmetic) or fixed percentage (geometric) before choosing your formula.

Question 14

A company deposits money into an account at the end of each month. The first deposit is $200, and each month's deposit is 5%5\% larger than the previous month's deposit. What is the total amount deposited after 88 months (ignore interest earned on the account itself)?

  1. 200(1−1.058)1−1.05\dfrac{200\big(1-1.05^8\big)}{1-1.05} (correct answer)
  2. 200(1−1.057)1−1.05\dfrac{200\big(1-1.05^7\big)}{1-1.05}
  3. 200⋅8⋅1.05200\cdot 8\cdot 1.05
  4. 200(1−0.958)1−0.95\dfrac{200\big(1-0.95^8\big)}{1-0.95}
Explanation: This question tests your understanding of geometric series—the sum of terms from a geometric sequence—and applying the formula Sn=a1−rn1−rS_n = a \frac{1 - r^n}{1 - r} to real-world growth scenarios like deposits. A geometric series is the sum of terms from a geometric sequence: here deposits are 200,200×1.05,200×1.052,…200, 200 \times 1.05, 200 \times 1.05^2, \dots, for 8 months (multiplying by 1.05 each time). Rather than listing and adding, use the formula with a=200, r=1.05, n=8. The total is S8=2001−1.0581−1.05S_8 = 200 \frac{1 - 1.05^8}{1 - 1.05}; note the negative denominator, but it yields positive sum. Choice A correctly identifies parameters and applies the formula accurately for n=8 terms. Choice B uses n=7 with r^7, perhaps undercounting terms—remember, first deposit is term 1, up to 8th! The three-step strategy: (1) Spot the geometric pattern in the scenario, identify a, r, n, (2) Confirm r ≠1, (3) Plug in and compute. Modeling growth with geometrics is super useful in finance—nice work exploring it!

Question 15

A ball is dropped from a height of 120 feet. Each time it bounces, it reaches 60% of its previous height. What is the total vertical distance traveled by the ball after it completes 6 bounces (including the initial drop)?

  1. 264.48264.48 feet
  2. 288.96288.96 feet (correct answer)
  3. 312.24312.24 feet
  4. 240.72240.72 feet
Explanation: The ball travels down 120 feet initially. Then it bounces up and down: up 120(0.6)=72120(0.6) = 72, down 72, up 120(0.6)2=43.2120(0.6)^2 = 43.2, down 43.2, etc. Total distance = 120+2[72+43.2+25.92+15.55+9.33+5.60]120 + 2[72 + 43.2 + 25.92 + 15.55 + 9.33 + 5.60]. The sum in brackets is a geometric series: 72+120(0.6)2+⋯+120(0.6)6=120(0.6)[1+0.6+(0.6)2+⋯+(0.6)5]=72⋅1−(0.6)61−0.6=72⋅1−0.0466560.4=72⋅2.3834=171.672 + 120(0.6)^2 + \cdots + 120(0.6)^6 = 120(0.6)[1 + 0.6 + (0.6)^2 + \cdots + (0.6)^5] = 72 \cdot \frac{1-(0.6)^6}{1-0.6} = 72 \cdot \frac{1-0.046656}{0.4} = 72 \cdot 2.3834 = 171.6. But we need the up distances: 120(0.6)⋅1−(0.6)61−0.6=84.48120(0.6) \cdot \frac{1-(0.6)^6}{1-0.6} = 84.48. Total = 120+2(84.48)=288.96120 + 2(84.48) = 288.96 feet. Choice A omits one bounce, C adds an extra bounce, D counts only up distances.

Question 16

Derive the finite geometric series sum formula. Let Sn=a+ar+ar2+⋯+arn−1S_n = a + ar + ar^2 + \cdots + ar^{n-1} with r≠1r \ne 1. Multiply by rr and subtract to show the telescoping cancellation, then solve for SnS_n. Which expression is correct?

  1. Sn=a(1−rn)1+rS_n = \dfrac{a(1-r^n)}{1+r}
  2. Sn=a(1−rn)1−rS_n = \dfrac{a(1-r^n)}{1-r} (correct answer)
  3. Sn=a+nrS_n = a + nr
  4. Sn=a(1−rn−1)1−rS_n = \dfrac{a(1-r^{n-1})}{1-r}
Explanation: This question tests your understanding of geometric series—the sum of terms from a geometric sequence—and how to derive the formula Sn=a(1−rn)/(1−r)S_n = a(1 - r^n) / (1 - r) through telescoping cancellation for efficient summation. A geometric series is Sn=a+ar+ar2+⋯+arn−1S_n = a + ar + ar^2 + \dots + ar^{n-1}, and multiplying by r gives rSn=ar+ar2+⋯+arnr S_n = ar + ar^2 + \dots + ar^n; subtracting them cancels intermediate terms. Rather than just memorizing, deriving it step-by-step builds deep insight: Sn−rSn=a−arnS_n - r S_n = a - ar^n, so Sn(1−r)=a(1−rn)S_n (1 - r) = a(1 - r^n), then divide to get the formula. This telescoping is the key—most terms vanish, leaving a simple expression! Choice B correctly shows the derived formula with exponent n and denominator 1-r. Choice A uses n-1 in the exponent, which would be off by one term, tempting if you miscount the series starting at ar^0; remember the last term is ar^{n-1}, so r^n in the formula. Master the derivation by practicing the subtraction on paper, check r≠1r \ne 1 (use n a if so), and apply to small series to verify—this process empowers you to handle any geometric sum confidently!

Question 17

Consider the finite geometric series ∑k=094(13)k.\sum_{k=0}^{9} 4\left(\frac{1}{3}\right)^k. What is its exact value? Note: Since ∣13∣<1\left|\frac{1}{3}\right|<1, the infinite geometric series would converge, but this question asks for the finite sum with n=10n=10 terms.

  1. 4(13)104\left(\frac{1}{3}\right)^{10}
  2. 4(1−(13)9)1−13\dfrac{4\left(1-\left(\frac{1}{3}\right)^9\right)}{1-\frac{1}{3}}
  3. 4(1−(13)10)1−13\dfrac{4\left(1-\left(\frac{1}{3}\right)^{10}\right)}{1-\frac{1}{3}} (correct answer)
  4. 4(1−(13)10)1+13\dfrac{4\left(1-\left(\frac{1}{3}\right)^{10}\right)}{1+\frac{1}{3}}
Explanation: This question tests your understanding of geometric series—the sum of terms from a geometric sequence—and applying Sn=a(1−rn)/(1−r)S_n = a(1 - r^n) / (1 - r) to summation notation with ∣r∣<1|r|<1. The sum ∑k=094(1/3)k\sum_{k=0}^9 4 (1/3)^k = 4∑k=09(1/3)k4 \sum_{k=0}^9 (1/3)^k, with inner sum (1−(1/3)10)/(1−1/3)(1 - (1/3)^{10}) / (1 - 1/3), n=10 terms. Rather than expanding, the formula handles it; note for infinite, it converges to a/(1−r)a/(1-r), but here finite. Derivation's telescoping works for any n! Choice A correctly uses exponent 10 and denominator 1 - 1/3. Choice B has 9, perhaps off-by-one in indexing—k=0 to 9 is 10 terms. Confirm range (10 terms), simplify denominator (1 - 1/3 = 2/3), and compare to infinite for insight—this finite vs. infinite distinction is key, you're doing wonderfully!

Question 18

A savings plan deposits $200 at the end of each month into an account that earns 1%1\% interest per month. If the account starts at $0, what is the total amount in the account immediately after the 6th deposit (ignore any fees)?

Model the balance as a finite geometric series: 200(1+1.01+1.012+⋯+1.015).200\big(1 + 1.01 + 1.01^2 + \cdots + 1.01^5\big).

  1. 200⋅1−1.0161−1.01200\cdot \dfrac{1-1.01^6}{1-1.01} (correct answer)
  2. 200⋅1−1.0151−1.01200\cdot \dfrac{1-1.01^5}{1-1.01}
  3. 200⋅1.016−11−1.01200\cdot \dfrac{1.01^6-1}{1-1.01}
  4. 200⋅1.016200\cdot 1.01^6
Explanation: This question tests your understanding of geometric series—the sum of terms from a geometric sequence—and applying the formula Sn=a(1−rn)/(1−r)S_n = a(1 - r^n) / (1 - r) to model real-world scenarios like savings with interest. The balance after 6 deposits is 200(1+1.01+1.012+⋯+1.015)200(1 + 1.01 + 1.01^2 + \cdots + 1.01^5), where each term represents a deposit's growth, with a=200a=200. Actually, it's 200 times sum of geometric terms with first=1, r=1.01, n=6. Rather than listing each, the formula captures the pattern: sum=(1−1.016)/(1−1.01)sum = (1 - 1.01^6)/(1 - 1.01), then multiply by 200. Derivation shows terms cancel nicely, perfect for compounding! Choice A correctly uses exponent 6 for n=6 terms and denominator 1−1.011-1.01. Choice C flips numerator to 1.016−11.01^6 -1 but keeps 1−1.011-1.01, making it negative—remember equivalent form is (rn−1)/(r−1)(r^n -1)/(r-1) for positive result. Identify the series structure (here starting at 1, not 200 as aa), count terms accurately, and simulate with small n to verify—this modeling skill is powerful for finance, you've got it!

Question 19

Use the geometric series sum formula to find the sum of the first 6 terms of the series 3+6+12+24+48+96.3+6+12+24+48+96.

  1. 9696
  2. 189189 (correct answer)
  3. 195195
  4. 192192
Explanation: This question tests your understanding of geometric series—the sum of terms from a geometric sequence—and how to derive and use the formula Sn=a1−rn1−rS_n = a \frac{1 - r^n}{1 - r} to calculate these sums efficiently. A geometric series is the sum of terms from a geometric sequence: here we have 3, 6, 12, 24, 48, 96 where each term is multiplied by 2 to get the next. Rather than adding manually (3+6+12+24+48+963 + 6 + 12 + 24 + 48 + 96), we use the formula Sn=a1−rn1−rS_n = a \frac{1 - r^n}{1 - r}, where a is first term, r is common ratio, and n is number of terms being summed. To apply the formula: identify a = 3 (first term), r = 2 (each term is 2 times the previous: 6/3 = 2, 12/6 = 2, etc.), and n = 6 (we're summing 6 terms). Substitute into S6=31−261−2=31−64−1=3−63−1=189S_6 = 3 \frac{1 - 2^6}{1 - 2} = 3 \frac{1 - 64}{-1} = 3 \frac{-63}{-1} = 189. Choice A correctly gives 189 as the sum. Choice B gives 192, which might come from miscounting terms or calculation error—always double-check your arithmetic, especially with negative signs! The three-step geometric series strategy: (1) Identify a, r, n by examining the series carefully. (2) Check r not equal 1 (here r = 2, so we're good). (3) Substitute into Sn=a1−rn1−rS_n = a \frac{1 - r^n}{1 - r} and calculate step by step. Why this formula is powerful: manually adding 3 + 6 + 12 + 24 + 48 + 96 takes multiple operations and risks arithmetic errors, but the formula gives us 189 in just a few steps—that's mathematical efficiency!

Question 20

A ball is dropped from 10 ft. Each bounce reaches 80%80\% of the previous height. What is the total vertical distance traveled during the first 5 bounces (up-and-down for each bounce), not counting the initial drop?

Hint: The bounce heights form a geometric sequence with first bounce height 10(0.8)10(0.8).

  1. 20(10⋅0.8)1−0.851−0.820\left(10\cdot 0.8\right)\dfrac{1-0.8^5}{1-0.8}
  2. 2(10⋅0.8)1−0.851−0.8=53.7856 ft2\left(10\cdot 0.8\right)\dfrac{1-0.8^5}{1-0.8}=53.7856\text{ ft} (correct answer)
  3. 2(10⋅0.8)1−0.841−0.8=49.7328 ft2\left(10\cdot 0.8\right)\dfrac{1-0.8^4}{1-0.8}=49.7328\text{ ft}
  4. 2(10)1−0.851−0.8=67.232 ft2\left(10\right)\dfrac{1-0.8^5}{1-0.8}=67.232\text{ ft}
Explanation: This question tests your understanding of geometric series—the sum of terms from a geometric sequence—and how to derive and use the formula S sub n = a(1 - r to the n) divided by (1 - r) to calculate these sums efficiently. A bouncing ball problem creates a geometric series: first bounce reaches 10(0.8) = 8 ft, second reaches 8(0.8) = 6.4 ft, and so on. Since each bounce goes up AND down, we count distance twice per bounce: total distance = 2[8 + 6.4 + 5.12 + ...] for 5 bounces. To find the sum of bounce heights: a = 10(0.8) = 8 (first bounce height), r = 0.8 (each bounce is 80% of previous), n = 5 (five bounces). Using the formula: S sub 5 = 8(1 - 0.8 to the 5) divided by (1 - 0.8) = 8(1 - 0.32768) divided by 0.2 = 8(0.67232) divided by 0.2 = 26.8928 ft. Since each bounce goes up and down, total distance = 2 × 26.8928 = 53.7856 ft. Choice B correctly calculates this as 2(10 × 0.8)(1 - 0.8 to the 5) divided by (1 - 0.8) = 53.7856 ft. Choice D forgets that the first bounce height is 10(0.8) = 8, not 10—the ball doesn't bounce back to its original height! The three-step strategy for physics applications: (1) Identify what creates the geometric sequence (here, each bounce is 80% of previous). (2) Determine first term carefully (first bounce, not initial height). (3) Account for all motion (up AND down for bounces). Real-world geometric series often have these subtleties!