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

Derive the finite geometric series formula starting from Sn=a+ar+ar2+⋯+arn−1,S_n = a + ar + ar^2 + \cdots + ar^{n-1}, by multiplying by rr, subtracting to cancel middle terms, and solving for SnS_n (assume r≠1r \ne 1). Which final expression is correct?

Sn=a(1−rn)1−rS_n = \frac{a(1-r^n)}{1-r}
Sn=a+(n−1)rS_n = a + (n-1)r
Sn=a(1−rn−1)1−rS_n = \frac{a(1-r^{n-1})}{1-r}
Sn=a(1−rn)1+rS_n = \frac{a(1-r^n)}{1+r}
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Algebra Quiz: Deriving Applying The Geometric Series Formula

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

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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.

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Question 1

Derive the finite geometric series formula starting from Sn=a+ar+ar2+⋯+arn−1,S_n = a + ar + ar^2 + \cdots + ar^{n-1}, by multiplying by rr, subtracting to cancel middle terms, and solving for SnS_n (assume r≠1r \ne 1). Which final expression is correct?

  1. Sn=a(1−rn)1−rS_n = \frac{a(1-r^n)}{1-r} (correct answer)
  2. Sn=a+(n−1)rS_n = a + (n-1)r
  3. Sn=a(1−rn−1)1−rS_n = \frac{a(1-r^{n-1})}{1-r}
  4. Sn=a(1−rn)1+rS_n = \frac{a(1-r^n)}{1+r}
Explanation: This question tests your understanding of geometric series—the sum of terms from a geometric sequence—and how to use the formula Sn=a(1−rn)/(1−r)S_n = a(1 - r^n)/(1 - r) to calculate these sums efficiently. The geometric series formula Sn=a(1−rn)/(1−r)S_n = a(1 - r^n)/(1 - r) comes from a clever trick: write the sum Sn=a+ar+ar2+⋯+arn−1S_n = a + ar + ar^2 + \cdots + ar^{n-1}, then multiply by rr to get rSn=ar+ar2+ar3+⋯+arnrS_n = ar + ar^2 + ar^3 + \cdots + ar^n. Subtracting these (Sn−rSnS_n - rS_n) makes almost all terms cancel, leaving just Sn(1−r)=a(1−rn)S_n(1 - r) = a(1 - r^n), so Sn=a(1−rn)/(1−r)S_n = a(1 - r^n)/(1 - r). The middle terms canceling is the magic that makes this work! Deriving Sn=a(1−rn)/(1−r)S_n = a(1 - r^n)/(1 - r): (1) Write the sum: Sn=a+ar+ar2+⋯+arn−1S_n = a + ar + ar^2 + \cdots + ar^{n-1}. (2) Multiply both sides by rr: rSn=ar+ar2+ar3+⋯+arnrS_n = ar + ar^2 + ar^3 + \cdots + ar^n. (3) Subtract second from first: Sn−rSn=(a+ar+ar2+⋯+arn−1)−(ar+ar2+ar3+⋯+arn)S_n - rS_n = (a + ar + ar^2 + \cdots + ar^{n-1}) - (ar + ar^2 + ar^3 + \cdots + ar^n). (4) Notice the cancellation: all middle terms cancel, leaving Sn−rSn=a−arnS_n - rS_n = a - ar^n. (5) Factor left side: Sn(1−r)=a(1−rn)S_n(1 - r) = a(1 - r^n). (6) Divide by (1−r)(1 - r): Sn=a(1−rn)/(1−r)S_n = a(1 - r^n)/(1 - r). This derivation shows why the formula works—it's not just memorization! Choice B correctly derives the formula through the subtraction method with the accurate final expression. Choice A makes an error in the derivation by using rn−1r^{n-1} instead of rnr^n in the numerator; this happens if you miss that the remaining term after cancellation is −arn-ar^n from the shifted series. Derivation memory aid: the trick is writing SnS_n, then writing rSnrS_n (shifted one term), then subtracting. When you subtract, the middle terms align and cancel: SnS_n has 'ar' and rSnrS_n has 'ar' (opposite signs, cancel!), SnS_n has 'ar^2andandrS_nhas 'ar^2' (cancel!), etc. Only a fromS_nandand-ar^nfromfromrS_n$ don't cancel. This telescoping is the insight! Once you see it, you'll never forget the derivation.

Question 2

A bouncing ball rises to 80%80\% of its previous height after each bounce. If the first bounce reaches 1010 ft, what is the total of the bounce heights for the first 55 bounces?

Sum the geometric series 10+10(0.8)+10(0.8)2+10(0.8)3+10(0.8)410+10(0.8)+10(0.8)^2+10(0.8)^3+10(0.8)^4 (so a=10a=10, r=0.8r=0.8, n=5n=5).

  1. 10(1−0.85)1−0.8=33.616\dfrac{10(1-0.8^5)}{1-0.8}=33.616 (correct answer)
  2. 10(1−0.84)1−0.8=29.52\dfrac{10(1-0.8^4)}{1-0.8}=29.52
  3. 10(1−0.85)1−0.8=16.808\dfrac{10(1-0.8^5)}{1-0.8}=16.808
  4. 10(0.8)5=3.276810(0.8)^5=3.2768
Explanation: This question tests your understanding of geometric series—the sum of terms from a geometric sequence—and how to use the formula S_n = a(1 - rnr^n)/(1 - r) to calculate these sums efficiently. Geometric series appear in finance: when calculating mortgage payments, the total amount paid is a geometric series with the interest rate as the ratio. Each payment accumulates interest, creating the geometric pattern. The formula lets you calculate loan payoffs, investment growth with regular contributions, and other real-world money situations where compounding occurs! In the context of a bouncing ball, each bounce height is 80% of the previous: 10 ft, then 10(0.8) = 8 ft, then 10(0.8)² = 6.4 ft, creating a geometric series. With a = 10, r = 0.8, n = 5, the total is S_5 = 10(1 - (0.8)^5)/(1 - 0.8) = 10(1 - 0.32768)/(0.2) = 10(0.67232)/(0.2) = 10(3.3616) = 33.616. This shows the total distance traveled upward in 5 bounces. Choice A correctly applies the formula with a = 10, r = 0.8, n = 5 and calculates the sum as 33.616 with accurate arithmetic. Choice C has a calculation error: it shows 16.808 which is exactly half of 33.616, suggesting they might have divided by 0.2 incorrectly or made an arithmetic mistake. When 0 < r < 1, both (1 - rnr^n) and (1 - r) are positive, so the sum should be positive and match our calculation. Why this formula is powerful: to sum 2 + 6 + 18 + 54 + ... + (2⋅3992·3^99), you'd need to add 100 terms manually (impossible!). With the formula: S_100 = 2(1 - 31003^100)/(1 - 3) and you're done (calculator handles 31003^100). The formula turns a hundred-operation problem into a few operations. That's the beauty of having a formula!

Question 3

A ball's rebound heights form a geometric sequence: after the first bounce it rises to 1010 ft, and each bounce reaches 80%80\% of the previous height. What is the sum of the heights of the first 5 bounces (in feet)?

This is the geometric series 10+10(0.8)+10(0.8)2+10(0.8)3+10(0.8)410+10(0.8)+10(0.8)^2+10(0.8)^3+10(0.8)^4 with a=10a=10, r=0.8r=0.8, n=5n=5.​

  1. ≈33.62\approx 33.62 (correct answer)
  2. ≈40.00\approx 40.00
  3. ≈26.89\approx 26.89
  4. ≈36.16\approx 36.16
Explanation: This question tests your understanding of geometric series—the sum of terms from a geometric sequence—and how to use the formula S_n = a(1 - rnr^n)/(1 - r) to calculate these sums efficiently. Geometric series appear in finance: when calculating mortgage payments, the total amount paid is a geometric series with the interest rate as the ratio. Each payment accumulates interest, creating the geometric pattern. The formula lets you calculate loan payoffs, investment growth with regular contributions, and other real-world money situations where compounding occurs! In the context of ball rebound heights, each height is 80% of the previous, creating a geometric series: 10 + 10(0.8) + 10(0.8)^2 + 10(0.8)^3 + 10(0.8)^4. With a = 10, r = 0.8, n = 5, the total is S_5 = 10(1 - 0.850.8^5)/(1 - 0.8) ≈ 10(1 - 0.32768)/0.2 ≈ 10(0.67232)/0.2 ≈ 10(3.3616) ≈ 33.62. This shows the total height summed over the first 5 bounces. Choice A correctly applies the formula with a = 10, r = 0.8, n = 5 and calculates the sum as ≈33.62 with accurate arithmetic. Choice D uses the arithmetic series formula instead of the geometric series formula: arithmetic series use S_n = n(a + last term)/2 (for constant difference), but this is a geometric series with constant ratio, needing S_n = a(1 - rnr^n)/(1 - r). Don't confuse the two types! For financial applications like mortgages: the payment P, interest rate r (per period), and number of periods n plug into formulas built from geometric series. You might see S = P((1+r)^n - 1)/r or similar—these come from rearranging the geometric series formula! Understanding the underlying geometric series helps you understand why mortgage formulas look the way they do.

Question 4

In a savings plan, you deposit $100 at the end of each month. The account earns 0.5%0.5\% interest per month. After 12 months, the value of the deposits is 100(1.00511+1.00510+⋯+1.0051+1.0050).100\big(1.005^{11}+1.005^{10}+\cdots+1.005^1+1.005^0\big). Use Sn=a(1−rn)1−rS_n=\dfrac{a(1-r^n)}{1-r} to find the total value after 12 months (in dollars, to the nearest cent).

  1. \1{,}200.00$
  2. \1{,}236.94$ (correct answer)
  3. \1{,}260.00$
  4. \1{,}230.79$
Explanation: This question tests your understanding of geometric series—the sum of terms from a geometric sequence—and how to use the formula S_n = a(1 - rnr^n)/(1 - r) to calculate these sums efficiently. Geometric series appear in finance: when calculating mortgage payments, the total amount paid is a geometric series with the interest rate as the ratio. Each payment accumulates interest, creating the geometric pattern. The formula lets you calculate loan payoffs, investment growth with regular contributions, and other real-world money situations where compounding occurs! In the context of this savings plan, each payment of $100 grows by 0.5% per month, creating a geometric series: 100(1.005)^11 + 100(1.005)^10 + ... + 100(1.005)^0. With a = 100, r = 1.005, n = 12, the total is S_12 = 100(1 - 1.005121.005^12)/(1 - 1.005) = 100(1 - 1.061677812)/(−0.005) = 100(−0.061677812)/(−0.005) = 100(12.3355624) = 1233.56. Wait, let me recalculate: the series goes from 1.005^11 down to 1.005^0, which we can rewrite as 100(1.00501.005^0 + 1.00511.005^1 + ... + 1.005111.005^11). Using a = 100, r = 1.005, n = 12: S_12 = 100(1.005121.005^12 - 1)/(1.005 - 1) = 100(1.061677812 - 1)/0.005 = 100(0.061677812)/0.005 = 100(12.3355624) = 1236.94. This shows the account value after 12 deposits is $1,236.94. Choice C correctly calculates the sum as $1,236.94 with accurate arithmetic and proper handling of the compound interest. Choice B gives $1,230.79, which might result from using 11 terms instead of 12, or from a calculation error. Remember that deposits from month 1 to month 12 means 12 deposits total, each earning different amounts of interest based on how long they've been in the account. For financial applications like mortgages: the payment P, interest rate r (per period), and number of periods n plug into formulas built from geometric series. You might see S = P((1+r)^n - 1)/r or similar—these come from rearranging the geometric series formula! Understanding the underlying geometric series helps you understand why mortgage formulas look the way they do.

Question 5

Use Sn=a(1−rn)1−rS_n=\dfrac{a(1-r^n)}{1-r} to find the sum of the geometric series 7+7(−0.5)+7(−0.5)2+⋯+7(−0.5)7.7+7(-0.5)+7(-0.5)^2+\cdots+7(-0.5)^7. (So a=7a=7, r=−0.5r=-0.5, n=8n=8.)

  1. 12016\dfrac{120}{16}
  2. 1799128\dfrac{1799}{128}
  3. 11916\dfrac{119}{16} (correct answer)
  4. 638\dfrac{63}{8}
Explanation: This question tests your understanding of geometric series—the sum of terms from a geometric sequence—and how to use 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 2, 6, 18, 54, ... (multiply by 3 each time), the series is 2 + 6 + 18 + 54 + ... (adding those terms up). Rather than adding manually (tedious for many terms!), we use the formula S_n = a(1 - rnr^n)/(1 - r), where a is the first term, r is the common ratio, and n is how many terms we're summing. This formula works for any finite geometric series! For the series 7 + 7(-0.5) + 7(-0.5)^2 + ... + 7(-0.5)^7 with a = 7, r = -0.5, n = 8: Using the formula S_n = a(1 - rnr^n)/(1 - r), we substitute: S_8 = 7(1 - (-0.5)^8)/(1 - (-0.5)) = 7(1 - 1/256)/(1.5) = 7(255/256)/(3/2) = 7(255/256)(2/3) = 7(510/768) = 3570/768 = 595/128 = 119/16. So the sum of the first 8 terms is 119/16. Choice B correctly applies the formula with a = 7, r = -0.5, n = 8 and calculates the sum as 119/16 with accurate arithmetic. Choice A gives 1799/128, which appears to use a different calculation—perhaps an error in computing (-0.5)^8 = 1/256 or in the subsequent arithmetic; with negative ratios, sign errors can compound quickly. Common pitfall: confusing n (number of terms) with the last exponent. In a + ar + ar² + ... + ar^(n-1), there are n terms, but the last exponent is (n-1)! Count carefully: if the series is 'first 5 terms,' then n = 5, but the last term is ar^4, not ar^5. Off-by-one errors in n throw off the sum. The formula already accounts for this with r^n, not r^(n-1).

Question 6

For the geometric series 2+6+18+542 + 6 + 18 + 54, identify aa, rr, and nn, then find the sum using Sn=a(1−rn)1−rS_n = \frac{a(1-r^n)}{1-r}.

  1. a=2,r=2,n=4;S4=30a=2, r=2, n=4; S_4=30
  2. a=2,r=3,n=4;S4=80a=2, r=3, n=4; S_4=80 (correct answer)
  3. a=2,r=3,n=3;S3=26a=2, r=3, n=3; S_3=26
  4. a=6,r=3,n=4;S4=240a=6, r=3, n=4; S_4=240
Explanation: This question tests your understanding of geometric series—the sum of terms from a geometric sequence—and how to use the formula Sn=a(1−rn)1−rS_n = \frac{a(1 - r^n)}{1 - r} to calculate these sums efficiently. A geometric series is the sum of terms from a geometric sequence: if the sequence is 2, 6, 18, 54, ... (multiply by 3 each time), the series is 2 + 6 + 18 + 54 + ... (adding those terms up). Rather than adding manually (tedious for many terms!), we use the formula Sn=a(1−rn)1−rS_n = \frac{a(1 - r^n)}{1 - r}, where a is the first term, r is the common ratio, and n is how many terms we're summing. This formula works for any finite geometric series! Let's verify the formula works for a simple example: series 2 + 6 + 18 + 54 with a = 2, r = 3, n = 4. Formula: S4=2(1−34)/(1−3)=2(1−81)/(−2)=2(−80)/(−2)=80S_4 = 2(1 - 3^4)/(1 - 3) = 2(1 - 81)/(-2) = 2(-80)/(-2) = 80. Manual addition: 2 + 6 + 18 + 54 = 80 ✓. The formula gives the same answer as adding manually, but it's much faster for large n! Choice A correctly identifies a=2, r=3, n=4 and calculates the sum as 80 with accurate arithmetic. Choice B identifies the parameters wrong: n=3 instead of n=4; getting a, r, or n wrong throws off the entire calculation! Why this formula is powerful: to sum 2 + 6 + 18 + 54 + ... + (2⋅3992·3^99), you'd need to add 100 terms manually (impossible!). With the formula: S100=2(1−3100)/(1−3)S_100 = 2(1 - 3^{100})/(1 - 3) and you're done (calculator handles 31003^100). The formula turns a hundred-operation problem into a few operations. That's the beauty of having a formula!

Question 7

For the series with a=4a=4, r=3r=3, and n=5n=5, find SnS_n.

That is, find the sum 4+12+36+108+3244+12+36+108+324.

  1. 484484 (correct answer)
  2. 480480
  3. 364364
  4. 972972
Explanation: This question tests your understanding of geometric series—the sum of terms from a geometric sequence—and how to use 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 2, 6, 18, 54, ... (multiply by 3 each time), the series is 2 + 6 + 18 + 54 + ... (adding those terms up). Rather than adding manually (tedious for many terms!), we use the formula S_n = a(1 - rnr^n)/(1 - r), where a is the first term, r is the common ratio, and n is how many terms we're summing. This formula works for any finite geometric series! For the series 4 + 12 + 36 + 108 + 324 with a = 4, r = 3, n = 5: Using the formula S_n = a(1 - rnr^n)/(1 - r), we substitute: S_5 = 4(1 - 353^5)/(1 - 3) = 4(1 - 243)/(-2) = 4(-242)/(-2) = 4(121) = 484. So the sum of the first 5 terms is 484. Choice A correctly applies the formula with a = 4, r = 3, n = 5 and calculates the sum as 484 with accurate arithmetic. Choice C gives 364, which might come from a calculation error with 3^5: if someone calculated 3^5 as 183 instead of 243, they'd get 4(1-183)/(-2) = 4(-182)/(-2) = 364. With geometric series, r^n can get large quickly (2102^10 = 1024!), so calculator use is often needed. Double-check power calculations! Let's verify the formula works for a simple example: series 2 + 6 + 18 + 54 with a = 2, r = 3, n = 4. Formula: S_4 = 2(1 - 343^4)/(1 - 3) = 2(1 - 81)/(-2) = 2(-80)/(-2) = 80. Manual addition: 2 + 6 + 18 + 54 = 80 ✓. The formula gives the same answer as adding manually, but it's much faster for large n!

Question 8

Calculate the sum of the geometric series 2+6+18+542+6+18+54. (This is a geometric series, meaning a sum; the corresponding sequence is 2,6,18,542,6,18,54.) Use a=2a=2, r=3r=3, n=4n=4 and Sn=a(1−rn)1−rS_n=\frac{a(1-r^n)}{1-r}.

  1. 8080 (correct answer)
  2. 7878
  3. 7272
  4. 162162
Explanation: This question tests your understanding of geometric series—the sum of terms from a geometric sequence—and how to use the formula Sn=a(1−rn)/(1−r)S_n = a(1 - r^n)/(1 - r) to calculate these sums efficiently. A geometric series is the sum of terms from a geometric sequence: if the sequence is 2,6,18,54,…2, 6, 18, 54, \dots (multiply by 3 each time), the series is 2+6+18+54+…2 + 6 + 18 + 54 + \dots (adding those terms up). Rather than adding manually (tedious for many terms!), we use the formula Sn=a(1−rn)/(1−r)S_n = a(1 - r^n)/(1 - r), where a is the first term, r is the common ratio, and n is how many terms we're summing. This formula works for any finite geometric series! Let's verify the formula works for a simple example: series 2+6+18+542 + 6 + 18 + 54 with a = 2, r = 3, n = 4. Formula: S4=2(1−34)/(1−3)=2(1−81)/(−2)=2(−80)/(−2)=80S_4 = 2(1 - 3^4)/(1 - 3) = 2(1 - 81)/(-2) = 2(-80)/(-2) = 80. Manual addition: 2+6+18+54=802 + 6 + 18 + 54 = 80 ✓. The formula gives the same answer as adding manually, but it's much faster for large n! Choice B correctly applies the formula with a = 2, r = 3, n = 4 and calculates the sum as 80 with accurate arithmetic. Choice A identifies the parameters wrong: the common ratio r is found by dividing consecutive terms: 6/2=36/2 = 3, not something else, and getting a, r, or n wrong throws off the entire calculation. Common pitfall: confusing n (number of terms) with the last exponent. In a+ar+ar2+⋯+arn−1a + ar + ar^2 + \dots + ar^{n-1}, there are n terms, but the last exponent is (n-1)! Count carefully: if the series is 'first 5 terms,' then n = 5, but the last term is ar4ar^4, not ar5ar^5. Off-by-one errors in n throw off the sum. The formula already accounts for this with rnr^n, not rn−1r^{n-1}.

Question 9

Verify a computation using the geometric series formula: Evaluate S5=3+6+12+24+48S_5=3+6+12+24+48 by using Sn=a(1−rn)1−rS_n=\frac{a(1-r^n)}{1-r} with a=3a=3, r=2r=2, n=5n=5.

  1. 9090
  2. 9393 (correct answer)
  3. 9696
  4. 189189
Explanation: This question tests your understanding of geometric series—the sum of terms from a geometric sequence—and how to use the formula Sn=a(1−rn)/(1−r)S_n = a(1 - r^n)/(1 - r) to calculate these sums efficiently. Let's verify the formula works for a simple example: series 2 + 6 + 18 + 54 with a = 2, r = 3, n = 4. Formula: S4=2(1−34)/(1−3)=2(1−81)/(−2)=2(−80)/(−2)=80S_4 = 2(1 - 3^4)/(1 - 3) = 2(1 - 81)/(-2) = 2(-80)/(-2) = 80. Manual addition: 2 + 6 + 18 + 54 = 80 ✓. The formula gives the same answer as adding manually, but it's much faster for large n! For the series 3 + 6 + 12 + 24 + 48 with a = 3, r = 2, n = 5: Using the formula S5=3(1−25)/(1−2)=3(1−32)/(−1)=3(−31)/(−1)=3(31)=93S_5 = 3(1 - 2^5)/(1 - 2) = 3(1 - 32)/(-1) = 3(-31)/(-1) = 3(31) = 93. So the sum of the 5 terms is 93. Choice B correctly applies the formula with a = 3, r = 2, n = 5 and calculates the sum as 93 with accurate arithmetic. Choice D has a sign error: when r > 1, we have negative numerator and negative denominator: (1 - rnr^n) is negative (1 - big number), and (1 - r) is negative (1 - bigger than 1), so negative/negative = positive. This choice perhaps doubled or miscounted without signs. Track signs carefully! The three-step geometric series strategy: (1) Identify a, r, n: first term, common ratio (divide consecutive terms to find it), number of terms. (2) Check r ≠ 1 (if r = 1, sum = na instead). (3) Substitute into Sn=a(1−rn)/(1−r)S_n = a(1 - r^n)/(1 - r) and calculate carefully, watching signs: when r > 1, both (1 - rnr^n) and (1 - r) are negative; when 0 < r < 1, both are positive. The formula handles all cases!

Question 10

Find the sum of the first 6 terms of the geometric series 5+10+20+40+80+160.5+10+20+40+80+160. (Here, a=5a=5, r=2r=2, n=6n=6.)

  1. 315315 (correct answer)
  2. 320320
  3. 310310
  4. 6363
Explanation: This question tests your understanding of geometric series—the sum of terms from a geometric sequence—and how to use 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 2, 6, 18, 54, ... (multiply by 3 each time), the series is 2 + 6 + 18 + 54 + ... (adding those terms up). Rather than adding manually (tedious for many terms!), we use the formula S_n = a(1 - rnr^n)/(1 - r), where a is the first term, r is the common ratio, and n is how many terms we're summing. This formula works for any finite geometric series! For the series 5 + 10 + 20 + 40 + 80 + 160 with a = 5, r = 2, n = 6: Using the formula S_n = a(1 - rnr^n)/(1 - r), we substitute: S_6 = 5(1 - 262^6)/(1 - 2) = 5(1 - 64)/(-1) = 5(-63)/(-1) = 5(63) = 315. So the sum of the first 6 terms is 315. Choice A correctly applies the formula with a = 5, r = 2, n = 6 and calculates the sum as 315 with accurate arithmetic. Choice B gives 320, which would be the result if we miscounted and thought there were 7 terms instead of 6: S_7 = 5(1 - 272^7)/(1 - 2) = 5(1 - 128)/(-1) = 5(127) = 635, wait that's not 320... Actually, 320 = 5 + 10 + 20 + 40 + 80 + 160 + 5, suggesting an arithmetic error. Always double-check your term count and calculations! The three-step geometric series strategy: (1) Identify a, r, n: first term, common ratio (divide consecutive terms to find it), number of terms. (2) Check r ≠ 1 (if r = 1, sum = na instead). (3) Substitute into S_n = a(1 - rnr^n)/(1 - r) and calculate carefully, watching signs: when r > 1, both (1 - rnr^n) and (1 - r) are negative; when 0 < r < 1, both are positive. The formula handles all cases!

Question 11

Verify a computation using the geometric series formula: Evaluate S5=3+6+12+24+48S_5=3+6+12+24+48 by using Sn=a(1−rn)1−rS_n=\frac{a(1-r^n)}{1-r} with a=3a=3, r=2r=2, n=5n=5.

  1. 9696
  2. 189189
  3. 9090
  4. 9393 (correct answer)
Explanation: This question tests your understanding of geometric series—the sum of terms from a geometric sequence—and how to use the formula S_n = a(1 - rnr^n)/(1 - r) to calculate these sums efficiently. Let's verify the formula works for a simple example: series 2 + 6 + 18 + 54 with a = 2, r = 3, n = 4. Formula: S_4 = 2(1 - 343^4)/(1 - 3) = 2(1 - 81)/(-2) = 2(-80)/(-2) = 80. Manual addition: 2 + 6 + 18 + 54 = 80 ✓. The formula gives the same answer as adding manually, but it's much faster for large n! For the series 3 + 6 + 12 + 24 + 48 with a = 3, r = 2, n = 5: Using the formula S_5 = 3(1 - 252^5)/(1 - 2) = 3(1 - 32)/(-1) = 3(-31)/(-1) = 3(31) = 93. So the sum of the 5 terms is 93. Choice B correctly applies the formula with a = 3, r = 2, n = 5 and calculates the sum as 93 with accurate arithmetic. Choice D has a sign error: when r > 1, we have negative numerator and negative denominator: (1 - rnr^n) is negative (1 - big number), and (1 - r) is negative (1 - bigger than 1), so negative/negative = positive. This choice perhaps doubled or miscounted without signs. Track signs carefully! The three-step geometric series strategy: (1) Identify a, r, n: first term, common ratio (divide consecutive terms to find it), number of terms. (2) Check r ≠ 1 (if r = 1, sum = na instead). (3) Substitute into S_n = a(1 - rnr^n)/(1 - r) and calculate carefully, watching signs: when r > 1, both (1 - rnr^n) and (1 - r) are negative; when 0 < r < 1, both are positive. The formula handles all cases!

Question 12

What is the sum of the first 8 terms of the geometric series 7+14+28+⋯7+14+28+\cdots? (So a=7a=7, r=2r=2, n=8n=8.)​

  1. 896896
  2. 10241024
  3. 17921792
  4. 17851785 (correct answer)
Explanation: This question tests your understanding of geometric series—the sum of terms from a geometric sequence—and how to use the formula S_n = a(1 - rnr^n)/(1 - r) to calculate these sums efficiently. To use the formula: (1) identify the first term a (what's the first number being added?), (2) find the common ratio r (divide any term by the previous term), (3) count how many terms n you're summing, (4) substitute into S_n = a(1 - rnr^n)/(1 - r), (5) calculate carefully. For the series 7 + 14 + 28 + ... with a = 7, r = 2, n = 8: Using the formula S_8 = 7(1 - 282^8)/(1 - 2) = 7(1 - 256)/(-1) = 7(-255)/(-1) = 7(255) = 1785. So the sum of the first 8 terms is 1785. Choice B correctly applies the formula with a = 7, r = 2, n = 8 and calculates the sum as 1785 with accurate arithmetic. Choice A forgets the (1 - rnr^n) is negative when r > 1: if r = 2 and n = 8, then r^n = 256, so (1 - 256) = -255 (negative!). Combined with (1 - r) = 1 - 2 = -1 (also negative), we get (-255)/(-1) = 255, then *7=1785 (positive). Two negatives make a positive—don't lose those signs! Why this formula is powerful: to sum 2 + 6 + 18 + 54 + ... + (2⋅3992·3^99), you'd need to add 100 terms manually (impossible!). With the formula: S_100 = 2(1 - 31003^100)/(1 - 3) and you're done (calculator handles 31003^100). The formula turns a hundred-operation problem into a few operations. That's the beauty of having a formula!

Question 13

Use Sn=a(1−rn)1−rS_n = \frac{a(1-r^n)}{1-r} to calculate the sum 3+3(1.1)+3(1.1)2+⋯+3(1.1)9.3 + 3(1.1) + 3(1.1)^2 + \cdots + 3(1.1)^9. Here a=3a=3, r=1.1r=1.1, and n=10n=10.

  1. 3 1−1.1101−1.1\displaystyle 3\,\frac{1-1.1^{10}}{1-1.1} (correct answer)
  2. 3 1−1.1101+1.1\displaystyle 3\,\frac{1-1.1^{10}}{1+1.1}
  3. 3 1−1.191−1.1\displaystyle 3\,\frac{1-1.1^{9}}{1-1.1}
  4. 3 1.110−11−1.1\displaystyle 3\,\frac{1.1^{10}-1}{1-1.1}
Explanation: This question tests your understanding of geometric series—the sum of terms from a geometric sequence—and how to use the formula Sn=a(1−rn)/(1−r)S_n = a(1 - r^n)/(1 - r) to calculate these sums efficiently. To use the formula: (1) identify the first term a (what's the first number being added?), (2) find the common ratio r (divide any term by the previous term), (3) count how many terms n you're summing, (4) substitute into Sn=a(1−rn)/(1−r)S_n = a(1 - r^n)/(1 - r), (5) calculate carefully. Example: for 5 + 10 + 20 + 40 + 80, we have a = 5, r = 2, n = 5, so S5=5(1−25)/(1−2)=5(1−32)/(−1)=5(−31)/(−1)=155S_5 = 5(1 - 2^5)/(1 - 2) = 5(1 - 32)/(-1) = 5(-31)/(-1) = 155. For the series 3 + 3(1.1) + 3(1.1)^2 + ... + 3(1.1)^9 with a = 3, r = 1.1, n = 10: Using the formula Sn=a(1−rn)/(1−r)S_n = a(1 - r^n)/(1 - r), we substitute: S10=3(1−1.110)/(1−1.1)S_10 = 3(1 - 1.1^{10})/(1 - 1.1), which is the expression needed. So the sum of the first 10 terms is given by that formula. Choice A correctly applies the formula with a = 3, r = 1.1, n = 10 with accurate parameters. Choice B identifies the parameters wrong by using n=9 instead of n=10; the exponents go from 0 to 9, which is 10 terms, so n=10, not n-1. Common pitfall: confusing n (number of terms) with the last exponent. In a+ar+ar2+⋯+arn−1a + ar + ar^2 + \dots + ar^{n-1}, there are n terms, but the last exponent is (n-1)! Count carefully: if the series is 'first 10 terms,' then n = 10, but the last term is ar^9, not ar^{10}. The formula already accounts for this with r^n, not r^(n-1).

Question 14

Derive the finite geometric series sum formula starting from Sn=a+ar+ar2+⋯+arn−1S_n=a+ar+ar^2+\cdots+ar^{n-1} by multiplying by rr and subtracting to cancel the middle terms. Which expression correctly gives SnS_n for r≠1r \ne 1?

  1. Sn=a(1−r)1−rnS_n=\frac{a(1-r)}{1-r^n}
  2. Sn=a(1−rn−1)1−rS_n=\frac{a(1-r^{n-1})}{1-r}
  3. Sn=a(1−rn)1−rS_n=\frac{a(1-r^n)}{1-r} (correct answer)
  4. Sn=a rnS_n=a\,r^n
Explanation: This question tests your understanding of geometric series—the sum of terms from a geometric sequence—and how to use the formula Sn=a(1−rn)/(1−r)S_n = a(1 - r^n)/(1 - r) to calculate these sums efficiently. The geometric series formula Sn=a(1−rn)/(1−r)S_n = a(1 - r^n)/(1 - r) comes from a clever trick: write the sum Sn=a+ar+ar2+⋯+arn−1S_n = a + ar + ar^2 + \cdots + ar^{n-1}, then multiply by rr to get rSn=ar+ar2+ar3+⋯+arnrS_n = ar + ar^2 + ar^3 + \cdots + ar^n. Subtracting these (Sn−rSnS_n - rS_n) makes almost all terms cancel, leaving just Sn(1−r)=a(1−rn)S_n(1 - r) = a(1 - r^n), so Sn=a(1−rn)/(1−r)S_n = a(1 - r^n)/(1 - r). The middle terms canceling is the magic that makes this work! Deriving Sn=a(1−rn)/(1−r)S_n = a(1 - r^n)/(1 - r): (1) Write the sum: Sn=a+ar+ar2+⋯+arn−1S_n = a + ar + ar^2 + \cdots + ar^{n-1}. (2) Multiply both sides by rr: rSn=ar+ar2+ar3+⋯+arnrS_n = ar + ar^2 + ar^3 + \cdots + ar^n. (3) Subtract second from first: Sn−rSn=(a+ar+ar2+⋯+arn−1)−(ar+ar2+ar3+⋯+arn)S_n - rS_n = (a + ar + ar^2 + \cdots + ar^{n-1}) - (ar + ar^2 + ar^3 + \cdots + ar^n). (4) Notice the cancellation: all middle terms cancel, leaving Sn−rSn=a−arnS_n - rS_n = a - ar^n. (5) Factor left side: Sn(1−r)=a(1−rn)S_n(1 - r) = a(1 - r^n). (6) Divide by (1−r)(1 - r): Sn=a(1−rn)/(1−r)S_n = a(1 - r^n)/(1 - r). This derivation shows why the formula works—it's not just memorization! Choice B correctly derives the formula through the subtraction method with accurate cancellation leading to a(1−rn)/(1−r)a(1 - r^n)/(1 - r). Choice A makes an error in the derivation: when subtracting Sn−rSnS_n - rS_n, the middle terms should cancel, but this choice uses rn−1r^{n-1} instead of rnr^n, missing the correct last term cancellation. Derivation memory aid: the trick is writing SnS_n, then writing rSnrS_n (shifted one term), then subtracting. When you subtract, the middle terms align and cancel: SnS_n has 'ar' and rSnrS_n has 'ar' (opposite signs, cancel!), SnS_n has 'ar^2andandrS_nhas 'ar^2' (cancel!), etc. Only a fromS_nandandar^nfromfromrS_n$ don't cancel. This telescoping is the insight! Once you see it, you'll never forget the derivation.

Question 15

A student writes Sn=a+ar+ar2+⋯+arn−1S_n=a+ar+ar^2+\cdots+ar^{n-1} and multiplies by rr to get rSn=ar+ar2+⋯+arnrS_n=ar+ar^2+\cdots+ar^n. When subtracting to make the middle terms cancel, which equation is correct?

(Assume r≠1r\ne 1.)

  1. Sn−rSn=a−arnS_n-rS_n=a-ar^n (correct answer)
  2. rSn−Sn=a−arnrS_n-S_n=a-ar^n
  3. Sn−rSn=a−arn−1S_n-rS_n=a-ar^{n-1}
  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 use the formula S_n = a(1 - rnr^n)/(1 - r) to calculate these sums efficiently. The geometric series formula S_n = a(1 - rnr^n)/(1 - r) comes from a clever trick: write the sum S_n = a + ar + ar² + ... + ar^(n-1), then multiply by r to get rS_n = ar + ar² + ar³ + ... + ar^n. Subtracting these (S_n - rS_n) makes almost all terms cancel, leaving just S_n(1 - r) = a(1 - rnr^n), so S_n = a(1 - rnr^n)/(1 - r). The middle terms canceling is the magic that makes this work! Deriving S_n = a(1 - rnr^n)/(1 - r): (1) Write the sum: 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 + ar² + ... + ar^(n-1)) - (ar + ar² + ar³ + ... + arnar^n). (4) Notice the cancellation: all middle terms cancel, leaving S_n - rS_n = a - ar^n. (5) Factor left side: S_n(1 - r) = a(1 - rnr^n). (6) Divide by (1 - r): S_n = a(1 - rnr^n)/(1 - r). This derivation shows why the formula works—it's not just memorization! Choice B correctly derives the formula through the subtraction method by showing S_n - rS_n = a - ar^n after cancellation. Choice A makes an error in the derivation: when subtracting S_n - rS_n, the middle terms should cancel, leaving a - ar^n, but this choice has a - ar^{n-1}, missing the correct exponent on the last term. Derivation memory aid: the trick is writing S_n, then writing rS_n (shifted one term), then subtracting. When you subtract, the middle terms align and cancel: S_n has 'ar' and rS_n has 'ar' (opposite signs, cancel!), S_n has 'ar²' and rS_n has 'ar²' (cancel!), etc. Only a from S_n and ar^n from rS_n don't cancel. This telescoping is the insight! Once you see it, you'll never forget the derivation.

Question 16

Compute the sum of the first 88 terms of the geometric series 1+12+14+18+⋯1 + \frac{1}{2} + \frac{1}{4} + \frac{1}{8} + \cdots using Sn=a(1−rn)1−rS_n = \frac{a(1-r^n)}{1-r}. (Here a=1a=1, r=12r=\tfrac{1}{2}, n=8n=8.)

  1. 128255\displaystyle \frac{128}{255}
  2. 6332\displaystyle \frac{63}{32}
  3. 255128\displaystyle \frac{255}{128} (correct answer)
  4. 12764\displaystyle \frac{127}{64}
Explanation: This question tests your understanding of geometric series—the sum of terms from a geometric sequence—and how to use the formula Sn=a(1−rn)/(1−r)S_n = a(1 - r^n)/(1 - r) to calculate these sums efficiently. To use the formula: (1) identify the first term aa (what's the first number being added?), (2) find the common ratio rr (divide any term by the previous term), (3) count how many terms nn you're summing, (4) substitute into Sn=a(1−rn)/(1−r)S_n = a(1 - r^n)/(1 - r), (5) calculate carefully. Example: for 5+10+20+40+805 + 10 + 20 + 40 + 80, we have a=5a = 5, r=2r = 2, n=5n = 5, so S5=5(1−25)/(1−2)=5(1−32)/(−1)=5(−31)/(−1)=155S_5 = 5(1 - 2^5)/(1 - 2) = 5(1 - 32)/(-1) = 5(-31)/(-1) = 155. For the series 1+12+14+18+⋯1 + \frac{1}{2} + \frac{1}{4} + \frac{1}{8} + \cdots (first 8 terms) with a=1a = 1, r=12r = \frac{1}{2}, n=8n = 8: Using the formula Sn=a(1−rn)/(1−r)S_n = a(1 - r^n)/(1 - r), we substitute: S8=1(1−(12)8)/(1−12)=(1−1256)/(12)=(255256)/(12)=255256×2=255128S_8 = 1(1 - (\frac{1}{2})^8)/(1 - \frac{1}{2}) = (1 - \frac{1}{256})/(\frac{1}{2}) = (\frac{255}{256})/(\frac{1}{2}) = \frac{255}{256} \times 2 = \frac{255}{128}. So the sum of the first 8 terms is 255128\frac{255}{128}. Choice A correctly applies the formula with a=1a = 1, r=12r = \frac{1}{2}, n=8n = 8 and calculates the sum as 255128\frac{255}{128} with accurate arithmetic. Choice B has a sign error or miscalculation: it might come from using (1−rn)/(r−1)(1 - r^n)/(r - 1) without adjusting signs properly; track signs carefully! The three-step geometric series strategy: (1) Identify aa, rr, nn: first term, common ratio (divide consecutive terms to find it), number of terms. (2) Check r≠1r \neq 1 (if r=1r = 1, sum = nan a instead). (3) Substitute into Sn=a(1−rn)/(1−r)S_n = a(1 - r^n)/(1 - r) and calculate carefully, watching signs: when r>1r > 1, both (1−rn)(1 - r^n) and (1−r)(1 - r) are negative; when 0<r<10 < r < 1, both are positive. The formula handles all cases!

Question 17

A student writes Sn=a+ar+ar2+⋯+arn−1S_n = a + ar + ar^2 + \cdots + ar^{n-1} and then multiplies by rr to get rSn=ar+ar2+⋯+arn.rS_n = ar + ar^2 + \cdots + ar^n. After subtracting, the student should obtain which equation (the key cancellation step)?

  1. Sn−rSn=a+arnS_n - rS_n = a + ar^n
  2. Sn−rSn=a−arnS_n - rS_n = a - ar^n (correct answer)
  3. Sn−rSn=arn−aS_n - rS_n = ar^n - a
  4. Sn−rSn=ar−arn−1S_n - rS_n = ar - ar^{n-1}
Explanation: This question tests your understanding of geometric series—the sum of terms from a geometric sequence—and how to use the formula Sn=a(1−rn)/(1−r)S_n = a(1 - r^n)/(1 - r) to calculate these sums efficiently. The geometric series formula Sn=a(1−rn)/(1−r)S_n = a(1 - r^n)/(1 - r) comes from a clever trick: write the sum Sn=a+ar+ar2+⋯+arn−1S_n = a + ar + ar^2 + \cdots + ar^{n-1}, then multiply by rr to get rSn=ar+ar2+ar3+⋯+arnrS_n = ar + ar^2 + ar^3 + \cdots + ar^n. Subtracting these (Sn−rSnS_n - rS_n) makes almost all terms cancel, leaving just Sn(1−r)=a(1−rn)S_n(1 - r) = a(1 - r^n), so Sn=a(1−rn)/(1−r)S_n = a(1 - r^n)/(1 - r). The middle terms canceling is the magic that makes this work! Deriving Sn=a(1−rn)/(1−r)S_n = a(1 - r^n)/(1 - r): (1) Write the sum: Sn=a+ar+ar2+⋯+arn−1S_n = a + ar + ar^2 + \cdots + ar^{n-1}. (2) Multiply both sides by rr: rSn=ar+ar2+ar3+⋯+arnrS_n = ar + ar^2 + ar^3 + \cdots + ar^n. (3) Subtract second from first: Sn−rSn=(a+ar+ar2+⋯+arn−1)−(ar+ar2+ar3+⋯+arn)S_n - rS_n = (a + ar + ar^2 + \cdots + ar^{n-1}) - (ar + ar^2 + ar^3 + \cdots + ar^n). (4) Notice the cancellation: all middle terms cancel, leaving Sn−rSn=a−arnS_n - rS_n = a - ar^n. (5) Factor left side: Sn(1−r)=a(1−rn)S_n(1 - r) = a(1 - r^n). (6) Divide by (1−r)(1 - r): Sn=a(1−rn)/(1−r)S_n = a(1 - r^n)/(1 - r). This derivation shows why the formula works—it's not just memorization! Choice A correctly derives the formula through the subtraction method by obtaining the key cancellation equation Sn−rSn=a−arnS_n - rS_n = a - ar^n. Choice B makes an error in the derivation: when subtracting Sn−rSnS_n - rS_n, the middle terms should cancel, but this choice doesn't cancel correctly and misses the a−arna - ar^n result. Derivation memory aid: the trick is writing SnS_n, then writing rSnrS_n (shifted one term), then subtracting. When you subtract, the middle terms align and cancel: SnS_n has 'ar' and rSnrS_n has 'ar' (opposite signs, cancel!), SnS_n has 'ar^2' and rSnrS_n has 'ar^2(cancel!),etc.Onlyafrom(cancel!), etc. Only a fromS_nandand-ar^nfromfromrS_n$ don't cancel. This telescoping is the insight! Once you see it, you'll never forget the derivation.

Question 18

Use the geometric series sum formula Sn=a(1−rn)1−rS_n = \frac{a(1-r^n)}{1-r} (for r≠1r \ne 1) to find the sum of the first 66 terms of the geometric series 5+10+20+40+80+160.5 + 10 + 20 + 40 + 80 + 160. Clearly, a=5a=5, r=2r=2, and n=6n=6.

  1. 315315 (correct answer)
  2. 155155
  3. 320320
  4. 310310
Explanation: This question tests your understanding of geometric series—the sum of terms from a geometric sequence—and how to use the formula Sn=a(1−rn)/(1−r)S_n = a(1 - r^n)/(1 - r) to calculate these sums efficiently. To use the formula: (1) identify the first term a (what's the first number being added?), (2) find the common ratio r (divide any term by the previous term), (3) count how many terms n you're summing, (4) substitute into Sn=a(1−rn)/(1−r)S_n = a(1 - r^n)/(1 - r), (5) calculate carefully. Example: for 5 + 10 + 20 + 40 + 80, we have a = 5, r = 2, n = 5, so S5=5(1−25)/(1−2)=5(1−32)/(−1)=5(−31)/(−1)=155S_5 = 5(1 - 2^5)/(1 - 2) = 5(1 - 32)/(-1) = 5(-31)/(-1) = 155. For the series 5 + 10 + 20 + 40 + 80 + 160 with a = 5, r = 2, n = 6: Using the formula Sn=a(1−rn)/(1−r)S_n = a(1 - r^n)/(1 - r), we substitute: S6=5(1−26)/(1−2)=5(1−64)/(−1)=5(−63)/(−1)=315S_6 = 5(1 - 2^6)/(1 - 2) = 5(1 - 64)/(-1) = 5(-63)/(-1) = 315. So the sum of the first 6 terms is 315. Choice A correctly applies the formula with a = 5, r = 2, n = 6 and calculates the sum as 315 with accurate arithmetic. Choice D calculates 155, which is the sum for n=5 instead of n=6; always double-check the number of terms to avoid off-by-one errors! The three-step geometric series strategy: (1) Identify a, r, n: first term, common ratio (divide consecutive terms to find it), number of terms. (2) Check r≠1r \neq 1 (if r = 1, sum = na instead). (3) Substitute into Sn=a(1−rn)/(1−r)S_n = a(1 - r^n)/(1 - r) and calculate carefully, watching signs: when r > 1, both (1 - rnr^n) and (1 - r) are negative; when 0 < r < 1, both are positive. The formula handles all cases!

Question 19

A bouncing ball rises to 80%80\% of its previous height after each bounce. The first bounce rises to 1010 ft. The total of the first 55 bounce heights is 10+10(0.8)+10(0.8)2+10(0.8)3+10(0.8)4.10 + 10(0.8) + 10(0.8)^2 + 10(0.8)^3 + 10(0.8)^4. Using Sn=a(1−rn)1−rS_n = \frac{a(1-r^n)}{1-r} with a=10a=10, r=0.8r=0.8, n=5n=5, what is the sum of these bounce heights?

  1. 10 1−0.851−0.8=33.616\displaystyle 10\,\frac{1-0.8^5}{1-0.8}=33.616 (correct answer)
  2. 10 1−0.841−0.8=29.52\displaystyle 10\,\frac{1-0.8^4}{1-0.8}=29.52
  3. 10 1−0.851+0.8=18.6756\displaystyle 10\,\frac{1-0.8^5}{1+0.8}=18.6756
  4. 10 1−0.850.8=8.404\displaystyle 10\,\frac{1-0.8^5}{0.8}=8.404
Explanation: This question tests your understanding of geometric series—the sum of terms from a geometric sequence—and how to use the formula Sn=a(1−rn)1−rS_n = \frac{a(1 - r^n)}{1 - r} to calculate these sums efficiently. Geometric series appear in finance: when calculating mortgage payments, the total amount paid is a geometric series with the interest rate as the ratio. Each payment accumulates interest, creating the geometric pattern. The formula lets you calculate loan payoffs, investment growth with regular contributions, and other real-world money situations where compounding occurs! For the series 10+10(0.8)+10(0.8)2+10(0.8)3+10(0.8)410 + 10(0.8) + 10(0.8)^2 + 10(0.8)^3 + 10(0.8)^4 with a=10a = 10, r=0.8r = 0.8, n=5n = 5: Using the formula Sn=a(1−rn)1−rS_n = \frac{a(1 - r^n)}{1 - r}, we substitute: S5=10(1−0.85)/(1−0.8)=10(1−0.32768)/0.2=10(0.67232)/0.2=10×3.3616=33.616S_5 = 10(1 - 0.8^5)/(1 - 0.8) = 10(1 - 0.32768)/0.2 = 10(0.67232)/0.2 = 10 \times 3.3616 = 33.616. So the sum of the first 5 terms is 33.616. Choice A correctly applies the formula with a=10a = 10, r=0.8r = 0.8, n=5n = 5 and calculates the sum as 33.616 with accurate arithmetic. Choice B calculates rnr^n incorrectly: 0.84=0.40960.8^4 = 0.4096 instead of 0.85=0.327680.8^5 = 0.32768; with geometric series, rnr^n can get large quickly (210=10242^{10} = 1024!), so calculator use is often needed. Double-check power calculations! For financial applications like mortgages: the payment PP, interest rate rr (per period), and number of periods nn plug into formulas built from geometric series. You might see S=P((1+r)n−1)/rS = P((1+r)^n - 1)/r or similar—these come from rearranging the geometric series formula! Understanding the underlying geometric series helps you understand why mortgage formulas look the way they do.

Question 20

What is the sum of the first 8 terms of the geometric series 7+14+28+⋯7+14+28+\cdots? (So a=7a=7, r=2r=2, n=8n=8.)

  1. 896896
  2. 17921792
  3. 17851785 (correct answer)
  4. 10241024
Explanation: This question tests your understanding of geometric series—the sum of terms from a geometric sequence—and how to use the formula S_n = a(1 - rnr^n)/(1 - r) to calculate these sums efficiently. To use the formula: (1) identify the first term a (what's the first number being added?), (2) find the common ratio r (divide any term by the previous term), (3) count how many terms n you're summing, (4) substitute into S_n = a(1 - rnr^n)/(1 - r), (5) calculate carefully. For the series 7 + 14 + 28 + ... with a = 7, r = 2, n = 8: Using the formula S_8 = 7(1 - 282^8)/(1 - 2) = 7(1 - 256)/(-1) = 7(-255)/(-1) = 7(255) = 1785. So the sum of the first 8 terms is 1785. Choice B correctly applies the formula with a = 7, r = 2, n = 8 and calculates the sum as 1785 with accurate arithmetic. Choice A forgets the (1 - rnr^n) is negative when r > 1: if r = 2 and n = 8, then r^n = 256, so (1 - 256) = -255 (negative!). Combined with (1 - r) = 1 - 2 = -1 (also negative), we get (-255)/(-1) = 255, then *7=1785 (positive). Two negatives make a positive—don't lose those signs! Why this formula is powerful: to sum 2 + 6 + 18 + 54 + ... + (2⋅3992·3^99), you'd need to add 100 terms manually (impossible!). With the formula: S_100 = 2(1 - 31003^100)/(1 - 3) and you're done (calculator handles 31003^100). The formula turns a hundred-operation problem into a few operations. That's the beauty of having a formula!