Algebra 2 Quiz: Deriving Applying The Geometric Series Formula

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Deriving Applying The Geometric Series FormulaQuestion 1 of 20

A geometric series has first term $a=12$ , common ratio $r=-\tfrac{1}{3}$ , and $n=5$ terms. Find $S_5$ .

S_5=12\,\dfrac{1-(-1/3)^5}{1-(-1/3)}=\dfrac{181}{9}

S_5=12\,\dfrac{1-(-1/3)^4}{1-(-1/3)}=\dfrac{328}{27}

S_5=12\,\dfrac{1-(-1/3)^5}{1-(1/3)}=\dfrac{181}{6}

S_5=12\,\dfrac{1-(-1/3)^6}{1-(-1/3)}=\dfrac{544}{27}

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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

A geometric series has first term $a=12$ , common ratio $r=-\tfrac{1}{3}$ , and $n=5$ terms. Find $S_5$ .

$S_5=12\,\dfrac{1-(-1/3)^5}{1-(-1/3)}=\dfrac{181}{9}$ (correct answer)
$S_5=12\,\dfrac{1-(-1/3)^4}{1-(-1/3)}=\dfrac{328}{27}$
$S_5=12\,\dfrac{1-(-1/3)^5}{1-(1/3)}=\dfrac{181}{6}$
$S_5=12\,\dfrac{1-(-1/3)^6}{1-(-1/3)}=\dfrac{544}{27}$

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 geometric series with negative common ratio alternates signs: here with a = 12 and r = -1/3, we get 12, -4, 4/3, -4/9, 4/27, ... Rather than dealing with alternating fractions, we use the formula S sub n = a(1 - r to the n) divided by (1 - r). To find S sub 5: substitute a = 12, r = -1/3, n = 5. First calculate (-1/3) to the 5 = -1/243 (negative because odd power of negative number). Then S sub 5 = 12(1 - (-1/243)) divided by (1 - (-1/3)) = 12(1 + 1/243) divided by (1 + 1/3) = 12(244/243) divided by (4/3) = 12(244/243) × (3/4) = 181/9. Choice A correctly shows this calculation, though it could be simplified further. Choice B incorrectly uses n = 4 instead of n = 5—when finding S sub 5, use 5 terms! Also note that (-1/3) to the 4 = 1/81 (positive for even power), while (-1/3) to the 5 = -1/243 (negative for odd power). The three-step strategy for negative ratios: (1) Calculate r to the n carefully—negative base with odd exponent stays negative. (2) Handle signs in (1 - r to the n) and (1 - r) separately. (3) Simplify the final fraction. Negative ratios create alternating series, but the formula handles them perfectly!

Question 2

Find the sum of the first $6$ terms of the geometric series $3+6+12+24+48+96.$

$189$ (correct answer)
$192$
$93$
$186$

Explanation: This question tests your understanding of geometric series—the sum of terms from a geometric sequence—and how to apply the formula 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: 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 + 96), we use the formula with a = 3 (first term), r = 2 (common ratio), and n = 6 (number of terms). Substituting into S_6 = 3(1 -

2^6

)/(1 - 2) = 3(1 - 64)/(-1) = 3(-63)/(-1) = 189. Choice A correctly calculates this sum as 189. Choice B (192) might come from miscounting terms or arithmetic error, while C (93) appears to use only 5 terms instead of 6. The three-step geometric series strategy: (1) Identify a = 3, r = 2, n = 6, (2) Check r ≠ 1 (it's 2, so we're good), (3) Substitute carefully into the formula. Watch the signs: when r > 1, both (1 -

r^n

) and (1 - r) are negative, making the quotient positive!

Question 3

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

$\dfrac{2\big(1-(1.1)^9\big)}{1-1.1}$
$\dfrac{2\big(1-(1.1)^{10}\big)}{1-1.1}$ (correct answer)
$\dfrac{2\big(1-(1.1)^{10}\big)}{1+1.1}$
$2(1.1)^9$

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

S_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

S_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

S_{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

r^0

r^9

is 10 terms. (2) Identify a as the coefficient (here 2). (3) Apply the formula carefully. Remember: when terms go from

r^0

r^k

, you have (k+1) terms total!

Question 4

Verify a formula application: Which value equals the sum $3+6+12+24$ using $S_n=\dfrac{a(1-r^n)}{1-r}$ with $a=3$ , $r=2$ , $n=4$ ?

$\dfrac{3(1-2^3)}{1-2}=21$
$\dfrac{3(1-2^4)}{1-2}=45$ (correct answer)
$\dfrac{3(1-2^4)}{1-2}=24$
$\dfrac{3(1-2^5)}{1-2}=93$

Explanation: This question tests your understanding of geometric series—the sum of terms from a geometric sequence—and verifying formula application with S_n = a(1 -

r^n

)/(1 - r). A geometric series is the sum of terms from a geometric sequence: here 3 + 6 + 12 + 24 (multiply by 2, 4 terms). Rather than adding manually (3+6=9, +12=21, +24=45), use a=3, r=2, n=4: S_4 = 3(1 -

2^4

)/(1 - 2) = 3(1-16)/(-1) = 3(-15)/(-1) = 45. The formula matches the manual sum, building trust! Choice B correctly applies with n=4 and r^4, giving 45. Choice A uses n=3 with r^3, summing only first 3 terms (21)—count terms carefully! The three-step verification strategy: (1) Identify parameters, (2) Compute formula and manual sum, (3) Compare. This reinforces the formula's reliability—keep verifying!

Question 5

A ball is dropped from $10$ ft and each bounce reaches $80\%$ 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 $k$ is $2(10)(0.8^k)$ for $k=1,2,3,4,5$ .

$20\left(\dfrac{1-0.8^5}{1-0.8}\right)$
$16\left(\dfrac{1-0.8^5}{1-0.8}\right)$ (correct answer)
$20\left(\dfrac{1-0.8^6}{1-0.8}\right)$
$20\left(\dfrac{1-0.8^5}{1+0.8}\right)$

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 -

r^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 -

0.8^5

)/(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(

1-0.8^5

)/(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 6

Derive the finite geometric series sum formula. Start with $S_n=a+ar+ar^2+\cdots+ar^{n-1}.$ Multiply both sides by $r$ and subtract to show the telescoping cancellation. Which expression correctly results from subtracting $rS_n$ from $S_n$ ?

$S_n-rS_n=a-ar^n$ (correct answer)
$S_n-rS_n=a-ar^{n-1}$
$S_n-rS_n=ar-ar^{n}$
$S_n-rS_n=a+ar^n$

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 -

r^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 +

ar^2

ar^3

+ ... +

ar^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

-ar^n

). This telescoping cancellation is what makes the formula derivation work!

Question 7

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

$100\,\dfrac{1-(1.01)^6}{1-1.01}$ (correct answer)
$100\,\dfrac{1-(1.01)^5}{1-1.01}$
$100\,\dfrac{1-(1.01)^6}{1+1.01}$
$100\,\dfrac{1-(1.01)^7}{1-1.01}$

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

S_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 + \cdots + 100(1.01)^5

represents the value of all deposits at the time of the 6th deposit. To identify parameters:

a = 100

(first term, the 6th deposit made just now),

r = 1.01

(each earlier deposit has grown by factor 1.01 per month), and

n = 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:

S_6 = 100 \frac{1 - (1.01)^6}{1 - 1.01}

. Choice A correctly shows this expression. Choice B incorrectly uses

n = 5

—remember we're summing 6 deposits, even though the highest power is 5. The pattern is: 6th deposit (just made) =

100(1.01)^0

, 5th deposit =

100(1.01)^1

, ..., 1st deposit =

100(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

n

. Financial applications often have this "off by one" trap where

n

terms means highest power is (

n-1

Question 8

A finite geometric series has first term $a=12$ , common ratio $r=\tfrac{1}{3}$ , and $n=7$ terms. What is $S_7$ ? (Note: If $|r|<1$ , the infinite sum would converge, but here you must find the finite sum.)

$\dfrac{12\left(1-\left(\tfrac13\right)^7\right)}{1-\tfrac13}$ (correct answer)
$\dfrac{12\left(1-\left(\tfrac13\right)^6\right)}{1-\tfrac13}$
$\dfrac{12\left(1-\left(\tfrac13\right)^7\right)}{1+\tfrac13}$
$12\cdot 7$

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 9

A phone trade-in promotion gives you $400 today, but if you wait, the offer decreases by $15\%$ 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 $400 + 400(0.85) + 400(0.85)^2 + \cdots + 400(0.85)^5.$ What is the exact sum in geometric-series form?

$400\cdot \dfrac{1-0.85^5}{1-0.85}$
$400\cdot (1-0.85^6)$
$400\cdot \dfrac{1-0.85^6}{1+0.85}$
$400\cdot \dfrac{1-0.85^6}{1-0.85}$ (correct answer)

Explanation: This question tests your understanding of geometric series—the sum of terms from a geometric sequence—and using

S_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:

400 \frac{1 - 0.85^6}{1 - 0.85}

, capturing the decay. Cancellation in derivation makes it ideal for

|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 10

For the series $5+15+45+135+405,$ identify $a$ , $r$ , and $n$ , then find the sum $S_n$ .

$a=5,\ r=3,\ n=5;\ S_5=605$ (correct answer)
$a=15,\ r=3,\ n=5;\ S_5=605$
$a=5,\ r=5,\ n=3;\ S_3=125$
$a=5,\ r=3,\ n=4;\ S_4=200$

Explanation: This question tests your understanding of geometric series—identifying parameters and applying the formula S_n = a(1 -

r^n

)/(1 - r) to calculate sums efficiently. A geometric series 5 + 15 + 45 + 135 + 405 has first term a = 5, and checking the ratio: 15/5 = 3, 45/15 = 3, so common ratio r = 3. Counting the terms: we have 5 terms total, so n = 5. Using the formula: S_5 = 5(1 -

3^5

)/(1 - 3) = 5(1 - 243)/(-2) = 5(-242)/(-2) = 5(121) = 605. Choice A correctly identifies a = 5, r = 3, n = 5 and calculates S_5 = 605. Choice B incorrectly states a = 15 (that's the second term, not first!), while C has wrong values for both r and n, and D miscounts n = 4 instead of 5. The three-step strategy: (1) Identify first term a (what's written first), (2) Find r by dividing any term by the previous one, (3) Count the terms carefully for n. Remember: the first term is a, not ar!

Question 11

For the geometric series $5+15+45+135+405,$ identify $a$ , $r$ , and $n$ , then find $S_n$ .

$a=5,\ r=3,\ n=5,\ S_5=605$ (correct answer)
$a=5,\ r=3,\ n=5,\ S_5=600$
$a=15,\ r=3,\ n=5,\ S_5=605$
$a=5,\ r=5,\ n=3,\ S_3=125$

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 -

r^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 -

3^5

)/(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 12

A ball is dropped from $10$ ft and bounces to $80\%$ of its previous height each time. What is the total vertical distance traveled during the first $5$ 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.)

$\dfrac{10}{1-0.8}$
$20\left(0.8+0.8^2+0.8^3+0.8^4+0.8^5\right)$ (correct answer)
$10\left(1+0.8+0.8^2+0.8^3+0.8^4\right)$
$20\left(1+0.8+0.8^2+0.8^3+0.8^4\right)$

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 +

0.8^2

+ ... +

0.8^5

), 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 13

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

$S_n=\dfrac{a(1-r^{n-1})}{1-r}$
$S_n=\dfrac{a(1-r^n)}{1-r}$ (correct answer)
$S_n=\dfrac{a(1-r^n)}{1+r}$
$S_n=\dfrac{n(a+r)}{2}$

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 -

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 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 -

r^n

), (5) Divide by (1 - r): S_n = a(1 -

r^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 14

A savings plan deposits money at the end of each month. The first deposit is $100, and each month the deposit is multiplied by $1.05$ (a 5% increase). What is the total amount deposited after $6$ months (ignore interest on the account itself)?

$100\cdot\dfrac{1-(1.05)^6}{1-1.05}$ (correct answer)
$100\cdot\dfrac{1-(1.05)^5}{1-1.05}$
$100\cdot\dfrac{1-(1.05)^6}{1.05}$
$100(1.05)^6$

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 -

r^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 15

Use the finite geometric series formula $S_n=\dfrac{a(1-r^n)}{1-r}$ (for $r\ne 1$ ) to find the sum $2+2(1.1)+2(1.1)^2+\cdots+2(1.1)^9.$

$S_{10}=\dfrac{2(1.1^{10}-1)}{1-1.1}\approx 31.87$
$S_{10}=\dfrac{2(1-1.1^{10})}{1-1.1}\approx 31.87$ (correct answer)
$S_{10}=\dfrac{2(1-1.1^{9})}{1-1.1}\approx 28.97$
$S_{9}=\dfrac{2(1-1.1^{10})}{1-1.1}\approx 31.87$

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 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, which is 10 terms total (powers go from 0 to 9). Rather than adding manually, we use the formula S sub n = a(1 - r to the n) divided by (1 - r), where a = 2 (first term), r = 1.1 (common ratio), and n = 10 (number of terms from 1.1 to the 0 through 1.1 to the 9). Substituting: S sub 10 = 2(1 - 1.1 to the 10) divided by (1 - 1.1) = 2(1 - 2.5937...) divided by (-0.1) = 2(-1.5937...) divided by (-0.1) ≈ 31.87. Choice A correctly identifies all parameters: a = 2, r = 1.1, n = 10 (counting from 2(1.1) to the 0 to 2(1.1) to the 9 gives 10 terms), and applies the formula accurately. Choice B incorrectly uses 1.1 to the 9 in the formula instead of 1.1 to the 10, essentially calculating the sum of only 9 terms instead of 10—remember that n is the total number of terms being summed, not the highest power! The three-step geometric series strategy works perfectly here: (1) Identify a = 2, r = 1.1, n = 10 by counting powers from 0 to 9. (2) Check r ≠ 1 (r = 1.1, so formula applies). (3) Substitute and calculate carefully with your calculator. When counting terms with powers starting at 0, remember that 0 to 9 gives you 10 terms total, just like counting from 1 to 10!

Question 16

A geometric series is the sum of terms of a geometric sequence. Find the sum of the first $6$ terms of the geometric series $3+6+12+24+48+96.$ (Here $a=3$ , $r=2$ , $n=6$ .)

$189$ (correct answer)
$96$
$192$
$63$

Explanation: This question tests your understanding of geometric series—the sum of terms from a geometric sequence—and how to apply the formula

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: here the sequence is 3, 6, 12, 24, 48, 96 (multiply by 2 each time), the series is 3 + 6 + 12 + 24 + 48 + 96 (adding those terms). Rather than adding manually (tedious for many terms!), we use the formula

S_n = a(1 - r^n) / (1 - r)

, where a is first term, r is common ratio, and n is number of terms being summed. For this series, plug in a=3, r=2, n=6:

S_6 = 3(1 - 2^6)/(1 - 2) = 3(1 - 64)/(-1) = 3(-63)/(-1) = 189

; you can verify by adding step-by-step to build confidence. Choice A correctly applies the formula with accurate parameter identification and calculation, yielding 189. Choice C uses just the last term 96, perhaps forgetting to sum all terms—remember, the series is the sum, not a single term! The three-step geometric series strategy: (1) Identify a, r, n, (2) Check r ≠ 1 to avoid division by zero, (3) Substitute and calculate carefully, watching signs for r > 1. Understanding this formula turns tedious addition into quick computation—great job practicing!

Question 17

Verify the formula by computing both ways: Which value equals $S_4 = 3 + 6 + 12 + 24$ and also equals $\dfrac{3(1-2^4)}{1-2}$ ?

$42$
$45$ (correct answer)
$48$
$51$

Explanation: This question tests your understanding of geometric series—the sum of terms from a geometric sequence—and verifying the formula

S_n = a(1 - r^n) / (1 - r)

by comparing manual and formula sums. For 3 + 6 + 12 + 24,

a=3

r=2

n=4

, manual sum is

3+6=9

+12=21

+24=45

. Rather than just one method, equating both confirms the formula:

3(1-16)/(1-2)=3(-15)/(-1)=45

. The derivation's telescoping shows why it matches perfectly! Choice B correctly computes 45 from both ways. Choice A might come from

n=3

sum=21 doubled or error, but count terms: 4 here. Practice by computing small series both manually and with formula, watch for sign flips when

r>1

, and use this to build trust in the method— you're on a roll!

Question 18

Use the geometric series formula to verify the sum. Which expression equals $3+6+12+24?$

$\dfrac{3(1-2^4)}{1-2}$ (correct answer)
$\dfrac{3(1-2^4)}{2-1}$
$\dfrac{3(1-2^3)}{1-2}$
$\dfrac{6(1-2^4)}{1-2}$

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 geometric series is the sum of terms from a geometric sequence: here we have 3 + 6 + 12 + 24, where each term is multiplied by 2 to get the next (r = 2). Rather than adding manually (though 3 + 6 + 12 + 24 = 45 is easy to verify), we use the formula S sub n = a(1 - r to the n) divided by (1 - r), where a = 3 (first term), r = 2 (common ratio), and n = 4 (number of terms). Substituting: S sub 4 = 3(1 - 2 to the 4) divided by (1 - 2) = 3(1 - 16) divided by (-1) = 3(-15) divided by (-1) = 45. Choice A correctly applies the formula with a = 3, r = 2, n = 4, giving the expression 3(1 - 2⁴)/(1 - 2) which equals 45. Choice B incorrectly uses 2³ instead of 2⁴ in the numerator, which would give 3(1 - 8)/(-1) = 21, not the correct sum of 45—remember the exponent in the formula matches the number of terms! The three-step strategy confirms: (1) Identify a = 3, r = 2, n = 4. (2) Check r ≠ 1 (r = 2). (3) Apply formula to get 45. This verification problem shows how the formula gives the same result as direct addition, building confidence in the formula's reliability!

Question 19

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

$S_n = \dfrac{a(1 - r^n)}{1 + r}$
$S_n = \dfrac{a + (n-1)r}{2}$
$S_n = \dfrac{a(1 - r^n)}{1 - r}$ (correct answer)
$S_n = \dfrac{a(1 - r^{n-1})}{1 - r}$

Explanation: This question tests your understanding of geometric series—the sum of terms from a geometric sequence—and how to derive the formula

S_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, ar^2, \dots, ar^{n-1}

, the series is

S_n = a + ar + ar^2 + \dots + ar^{n-1}

. Rather than adding manually, we derive the formula by a clever trick: multiply by

r

and subtract to make terms cancel! To derive

S_n = \dfrac{a(1 - r^n)}{1 - r}

: (1) Write

S_n = a + ar + ar^2 + \dots + ar^{n-1}

, (2) Multiply by

r

r S_n = ar + ar^2 + \dots + ar^n

, (3) Subtract:

S_n - r S_n = a - ar^n

, (4) Factor:

S_n (1 - r) = a (1 - r^n)

, (5) Solve:

S_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

n

r^n

. Choice A makes an error in the derivation by using

r^{n-1}

instead of

r^n

: when subtracting, the last term is

-ar^n

, so

1 - r^n

, not

1 - r^{n-1}

! The three-step derivation strategy: (1) Write

S_n

and

r S_n

, (2) Subtract to cancel, (3) Solve for

S_n

. Why this is powerful: it handles any

n

without listing terms—math magic at work!

Question 20

For the geometric series $5+15+45+135+405,$ identify $a$ , $r$ , and $n$ , then find the sum $S_n$ .

$a=5,\ r=3,\ n=5;\ S_5=605$ (correct answer)
$a=5,\ r=3,\ n=4;\ S_4=200$
$a=5,\ r=5,\ n=5;\ S_5=3905$
$a=15,\ r=3,\ n=5;\ S_5=1815$

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 geometric series is the sum of terms from a geometric sequence: here we have 5 + 15 + 45 + 135 + 405, where each term is multiplied by 3 to get the next. Rather than adding manually, we identify parameters: a = 5 (first term), r = 15/5 = 3 (common ratio), n = 5 (counting the five terms), then use the formula S sub n = a(1 - r to the n) divided by (1 - r). Substituting: S sub 5 = 5(1 - 3 to the 5) divided by (1 - 3) = 5(1 - 243) divided by (-2) = 5(-242) divided by (-2) = 605. Choice A correctly identifies all parameters: a = 5 (first term), r = 3 (each term is triple the previous), n = 5 (five terms total), and calculates S sub 5 = 605. Choice B incorrectly counts only 4 terms instead of 5, while Choice D misidentifies the common ratio as 5 instead of 3—always verify r by dividing consecutive terms: 15/5 = 3, 45/15 = 3, etc. The three-step geometric series strategy: (1) Identify parameters by inspection—first term is what you see first, ratio is what you multiply by, count terms carefully. (2) Verify r ≠ 1 (here r = 3). (3) Apply formula with confidence. This series grows rapidly because r = 3 > 1, demonstrating exponential growth—the last term 405 is already 81 times the first term 5!