Deriving/Applying the Geometric Series Formula - Algebra 2

Verify the formula by computing both ways: Which value equals $S_4 = 3 + 6 + 12 + 24$ and also equals $\dfrac{3(1-2^4)}{1-2}$ ?
Answer: $45$ 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!
For the geometric series $5+15+45+135+405,$ identify $a$ , $r$ , and $n$ , then find $S_n$ .
Answer: $a=5,\ r=3,\ n=5,\ S_5=605$ 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. Looking at the series 5 + 15 + 45 + 135 + 405, we identify: a = 5 (first term), r = 3 (since 15/5 = 3, 45/15 = 3, etc.), and n = 5 (five terms total). Using the formula $S_n$ = a(1 - $r^n$ )/(1 - r): $S_5$ = 5(1 - $3^5$ )/(1 - 3) = 5(1 - 243)/(-2) = 5(-242)/(-2) = 5(121) = 605. Choice A correctly identifies all parameters and calculates the sum. Choice B uses n = 4 (miscounting the terms); choice C incorrectly identifies a = 15 (that's the second term, not the first); choice D mistakes r = 5 (but 15/5 = 3, not 5). The three-step strategy works perfectly: (1) Identify a = 5, r = 3, n = 5 by examining the pattern; (2) Verify r ≠ 1 (it's 3); (3) Apply the formula carefully. Remember: the first term is a, not the second term, and count all terms carefully!
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.)
Answer: $20\left(0.8+0.8^2+0.8^3+0.8^4+0.8^5\right)$ 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 2*8, 2*6.4, etc. (but exclude initial drop). Rather than listing, recognize total = 20 (0.8 + $0.8^2$ + ... + $0.8^5$ ), since 2*10*0.8 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!
Use the geometric series formula $S_n=\dfrac{a(1-r^n)}{1-r}$ (for $r\ne 1$ ) to find the sum $2+2(1.1)+2(1.1)^2+\cdots+2(1.1)^9.$
Answer: $\dfrac{2\big(1-(1.1)^{10}\big)}{1-1.1}$ 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 sums with decimal ratios. A geometric series is the sum of terms from a geometric sequence: here it's $2, 2*1.1, 2*1.1^2, \ldots, \text{ up to } 2*1.1^9$ (10 terms, multiplying by 1.1 each time). Rather than adding manually, use $S_n = a(1 - r^n)/(1 - r)$ , with $a=2$ , $r=1.1$ , $n=10$ . Notice the series is $2(1 + 1.1 + 1.1^2 + \ldots + 1.1^9)$ , but actually starts with $2=2*1.1^0$ , so full formula applies directly. Choice A correctly applies the formula with $a=2$ , $r=1.1$ , $n=10$ , giving $2(1 - 1.1^{10})/(1 - 1.1)$ . Choice C uses $1 + r$ in the denominator instead of $1 - r$ : that's a sign error, since for $r>1$ , denominator is negative, but the formula accounts for it! The three-step strategy: (1) Identify $a$ , $r$ , $n$ (count terms carefully), (2) Check $r \ne 1$ , (3) Substitute, handling decimals precisely. This formula shines for growth rates like interest—keep practicing, you've got this!
Use the geometric series sum formula $S_n = \dfrac{a(1-r^n)}{1-r}$ (for $r\ne 1$ ) to find the value of $2 + 2(1.1) + 2(1.1)^2 + \cdots + 2(1.1)^9.$
Answer: $\dfrac{2\big(1-1.1^{10}\big)}{1-1.1}$ 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 express sums without expanding all terms. Here, the series is 2 + 2(1.1) + 2 $(1.1)^2$ + ... + 2 $(1.1)^9$ , which factors to 2 times (1 + 1.1 + ... + $1.1^9$ ), so $a=2$ , $r=1.1$ , $n=10$ terms. Rather than adding manually (tedious for decimals!), we use $S_n = a(1 - r^n) / (1 - r)$ , directly matching the given expression with the exponent 10 and denominator 1-1.1. The formula comes from a clever algebraic trick where most terms cancel when you subtract $r S_n$ from $S_n$ —it's like magic that simplifies complex sums! Choice A correctly uses $n=10$ for the 10 terms (from exponent 0 to 9) and the standard denominator 1-r. Choice C flips the denominator to 1.1-1, which changes the sign and could lead to errors if not adjusted, but remember the standard form avoids confusion; Choice D omits the division, summing like a single term times $(r^n -1)$ , which overcounts. Always count terms carefully (here 10), verify $r>1$ handles negative denominators positively, and practice rewriting in equivalent forms like $(a(r^n -1))/(r-1)$ for flexibility—you're doing great, keep exploring these variations!
Find the sum of the first $6$ terms of the geometric series $3+6+12+24+48+96.$
Answer: $189$ 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!
Derive the finite geometric series sum formula. Let $S_n = a + ar + ar^2 + \cdots + ar^{n-1}$ with $r \ne 1$ . Multiply by $r$ and subtract to show the telescoping cancellation, then solve for $S_n$ . Which expression is correct?
Answer: $S_n = \dfrac{a(1-r^n)}{1-r}$ This question tests your understanding of geometric series—the sum of terms from a geometric sequence—and how to derive the formula $S_n = a(1 - r^n) / (1 - r)$ through telescoping cancellation for efficient summation. A geometric series is $S_n = a + ar + ar^2 + \dots + ar^{n-1}$ , and multiplying by r gives $r S_n = ar + ar^2 + \dots + ar^n$ ; subtracting them cancels intermediate terms. Rather than just memorizing, deriving it step-by-step builds deep insight: $S_n - r S_n = a - ar^n$ , so $S_n (1 - r) = a(1 - r^n)$ , 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 $r \ne 1$ (use n a if so), and apply to small series to verify—this process empowers you to handle any geometric sum confidently!
For the geometric series $5+15+45+135+405$ , identify $a$ , $r$ , and $n$ , then find $S_n$ (the sum of all terms shown).
Answer: $S_5=\dfrac{5(1-3^5)}{1-3}=605$ 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 5, 15, 45, 135, 405 where each term is multiplied by 3 to get the next. Rather than adding manually, we use the formula $S_n = a \frac{1 - r^n}{1 - r}$ , where $a$ is first term, $r$ is common ratio, and $n$ is number of terms being summed. To identify parameters: $a = 5$ (first term), $r = 3$ (each term is 3 times the previous: 15/5 = 3, 45/15 = 3, etc.), and $n = 5$ (we have 5 terms total). Substituting into the formula: $S_5 = 5 \frac{1 - 3^5}{1 - 3} = 5(1 - 243) / (-2) = 5(-242) / (-2) = 605$ . Choice A correctly identifies all parameters and calculates $S_5 = 605$ . Choice C incorrectly uses $n = 4$ instead of $n = 5$ —remember to count all terms shown! There are 5 terms (5, 15, 45, 135, 405), not 4. Choice D confuses the common ratio, using $r = 5$ instead of $r = 3$ . The three-step geometric series strategy: (1) Count terms carefully—don't skip the first or last! (2) Find $r$ by dividing any term by the previous one. (3) Apply the formula methodically. Understanding parameter identification prevents most errors in geometric series problems!
Compute the finite geometric sum $S_8=7+7\left(\tfrac12\right)+7\left(\tfrac12\right)^2+\cdots+7\left(\tfrac12\right)^7.$ (Note: This is finite; if $|r|<1$ , an infinite sum would converge, but here you are finding $S_8$ .)
Answer: $S_8=\dfrac{7\left(1-(\tfrac12)^8\right)}{1-\tfrac12}=\dfrac{1785}{128}$ 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 7 + 7(1/2) + 7(1/2) squared + ... + 7(1/2) to the 7, which is 8 terms total (powers go from 0 to 7). Rather than adding manually, we use the formula S sub n = a(1 - r to the n) divided by (1 - r), where a = 7 (first term), r = 1/2 (common ratio), and n = 8 (number of terms from (1/2) to the 0 through (1/2) to the 7). Substituting: S sub 8 = 7(1 - (1/2) to the 8) divided by (1 - 1/2) = 7(1 - 1/256) divided by (1/2) = 7(255/256) divided by (1/2) = 7(255/256) × 2 = 1785/128. Choice A correctly applies the formula with a = 7, r = 1/2, n = 8, recognizing that powers from 0 to 7 give 8 terms total. Choice B incorrectly uses n = 7 or calculates (1/2) to the 7 instead of (1/2) to the 8, essentially missing one term—remember to count all terms including the one with power 0! The three-step strategy works beautifully: (1) Identify a = 7, r = 1/2, n = 8 by counting powers. (2) Verify r ≠ 1 (r = 1/2). (3) Calculate carefully with fractions. When |r| < 1 like here, the terms get smaller and smaller—this finite sum is 1785/128 ≈ 13.95, and if we continued to infinity, it would approach 14!
Use the finite geometric series formula $S_n=\dfrac{a(1-r^n)}{1-r}$ (for $r\ne 1$ ) to find the sum $2+2(1.1)+2(1.1)^2+\cdots+2(1.1)^9.$
Answer: $S_{10}=\dfrac{2(1-1.1^{10})}{1-1.1}\approx 31.87$ 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!
Derive the finite geometric series sum formula. Start with $S_n=a+ar+ar^2+\cdots+ar^{n-1}$ for $r\ne 1$ . Multiply both sides by $r$ , subtract to show the middle terms cancel (telescoping), and solve for $S_n$ . Which expression is correct?
Answer: $S_n=\dfrac{a(1-r^n)}{1-r}$ 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) through the telescoping method. 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: $S_n$ - rS_n = (a + ar + ... + $ar^{n-1}$ ) - (ar + ar² + ... + $ar^n$ ), (4) Notice all middle terms cancel, leaving $S_n$ - rS_n = a - $ar^n$ , (5) Factor: $S_n$ (1 - r) = a(1 - $r^n$ ), (6) Divide: $S_n$ = a(1 - $r^n$ )/(1 - r). Choice B correctly shows the final formula with $r^n$ (not $r^{n-1}$ ) in the numerator, which comes from the last term $ar^n$ in the subtraction. Choice A incorrectly has $r^{n-1}$ , which would mean the series only went up to $ar^{n-2}$ ; Choice C inverts the formula structure; Choice D is the arithmetic series formula n(a+l)/2, not geometric. The key insight is recognizing that when you subtract rS_n from $S_n$ , the telescoping cancellation leaves only the first term of $S_n$ and the last term of rS_n!
For the series $5+15+45+135+405,$ identify $a$ , $r$ , and $n$ , then find the sum $S_n$ .
Answer: $a=5,\ r=3,\ n=5;\ S_5=605$ 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!
A savings plan deposits $200 at the end of each month into an account that earns $1\%$ 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: $200\big(1 + 1.01 + 1.01^2 + \cdots + 1.01^5\big).$
Answer: $200\cdot \dfrac{1-1.01^6}{1-1.01}$ This question tests your understanding of geometric series—the sum of terms from a geometric sequence—and applying the formula $S_n = a(1 - r^n) / (1 - r)$ to model real-world scenarios like savings with interest. The balance after 6 deposits is $200(1 + 1.01 + 1.01^2 + \cdots + 1.01^5)$ , where each term represents a deposit's growth, with $a=200$ . 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: $sum = (1 - 1.01^6)/(1 - 1.01)$ , then multiply by 200. Derivation shows terms cancel nicely, perfect for compounding! Choice A correctly uses exponent 6 for n=6 terms and denominator $1-1.01$ . Choice C flips numerator to $1.01^6 -1$ but keeps $1-1.01$ , making it negative—remember equivalent form is $(r^n -1)/(r-1)$ for positive result. Identify the series structure (here starting at 1, not 200 as $a$ ), count terms accurately, and simulate with small n to verify—this modeling skill is powerful for finance, you've got it!
Use the geometric series sum formula to find the sum of the first 6 terms of the series $3+6+12+24+48+96.$
Answer: $189$ 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 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 $S_n = a \frac{1 - r^n}{1 - r}$ , 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 $S_6 = 3 \frac{1 - 2^6}{1 - 2} = 3 \frac{1 - 64}{-1} = 3 \frac{-63}{-1} = 189$ . 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 $S_n = a \frac{1 - r^n}{1 - r}$ 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!
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?
Answer: $10,731.84 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 $a = 800$ , common ratio $r = 1.02$ , and $n = 12$ terms. The formula is: $S_n = a \cdot \frac{1-r^n}{1-r}$ 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) 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.
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?
Answer: $100\,\dfrac{1-(1.01)^6}{1-1.01}$ 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 ( $n-1$ )!
Find the sum of the first 8 terms of the geometric series with first term $a=12$ and common ratio $r=\tfrac{1}{2}$ . (Use $S_n=\dfrac{a(1-r^n)}{1-r}$ for $r\ne 1$ .)
Answer: $S_8=12\left(\dfrac{1-(1/2)^8}{1-1/2}\right)=\dfrac{765}{32}$ 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) when the common ratio is a fraction. A geometric series with 0 < r < 1 decreases term by term: starting with a = 12 and r = 1/2, the terms are 12, 6, 3, 1.5, 0.75, etc., getting smaller each time. For n = 8 terms, we calculate: $S_8$ = 12(1 - $(1/2)^8$ )/(1 - 1/2) = 12(1 - 1/256)/(1/2) = 12(255/256)/(1/2) = 12(255/256)(2/1) = 24(255/256) = 765/32. Choice A correctly applies the formula with proper fraction arithmetic: $(1/2)^8$ = 1/256, and dividing by (1 - 1/2) = 1/2 means multiplying by 2. Choice B uses n = 7 instead of 8; Choice C has the wrong denominator (should be 1 - 1/2 = 1/2, not just 1/2); Choice D uses the arithmetic series formula instead of geometric. The key with fractional ratios: the series converges (terms get smaller), and careful fraction arithmetic is essential—remember that dividing by 1/2 means multiplying by 2!
A store offers a rebate that pays $160 after the first month, and each month after that the rebate payment is multiplied by $0.75$ (so it decreases). The total rebate paid over 7 months is $160+160(0.75)+160(0.75)^2+\cdots+160(0.75)^6.$ What is the total rebate over 7 months, approximately?
Answer: $\approx \$424.69$ This question tests your understanding of geometric series—the sum of terms from a geometric sequence—and how to model decreasing payment scenarios using the formula $S_n$ = a(1 - $r^n$ )/(1 - r). A geometric series models situations where each term is a constant fraction of the previous: here, each month's rebate is 0.75 times (75% of) the previous month, creating the decreasing series 160 + 160(0.75) + 160(0.75)² + ... + 160(0.75)⁶. With a = 160, r = 0.75, n = 7 (months 1 through 7), we apply: $S_7$ = 160(1 - $0.75^7$ )/(1 - 0.75) = 160(1 - 0.1335)/(0.25) = 160(0.8665)/0.25 ≈ 424.69. Choice B correctly calculates this total rebate amount over 7 months. Choice A overestimates; Choice C underestimates significantly; Choice D ($600) might assume constant payments of about $85.71 per month. The key insight is that decreasing geometric series (0 < r < 1) converge to finite sums even with many terms—this models depreciation, cooling rates, and other decay processes. Understanding when to use geometric (multiplicative change) versus arithmetic (additive change) series is crucial for real-world modeling!
Compute the finite geometric series sum $1 - \frac{1}{2} + \frac{1}{4} - \frac{1}{8} + \frac{1}{16} - \frac{1}{32}.$
Answer: $\dfrac{21}{32}$ This question tests your understanding of geometric series—the sum of terms from a geometric sequence—and handling negative ratios in the formula $S_n = a \frac{1 - r^n}{1 - r}$ . A geometric series is the sum of terms from a geometric sequence: here $1, -\frac{1}{2}, \frac{1}{4}, -\frac{1}{8}, \frac{1}{16}, -\frac{1}{32}$ (multiply by $-\frac{1}{2}$ each time, 6 terms). Rather than adding manually, use $a=1$ , $r=-\frac{1}{2}$ , $n=6$ : $S_6 = 1 \frac{1 - (-\frac{1}{2})^6}{1 - (-\frac{1}{2})} = \frac{1 - \frac{1}{64}}{\frac{3}{2}} = \frac{63}{64} \times \frac{2}{3} = \frac{21}{32}.$ The alternating signs work because r is negative, but the formula handles it! Choice A correctly computes the sum using proper parameters, yielding $\frac{21}{32}$ . Choice C calculates $r^n$ incorrectly or miscounts terms: $(-\frac{1}{2})^6 = +\frac{1}{64}$ , not leading to $\frac{43}{64}$ —double-check powers! The three-step strategy: (1) Identify a, r (including sign), n, (2) Note formula works for $|r| <1$ or $>1$ , (3) Compute, simplifying fractions. Alternating series are fun—excellent effort!
A loan is repaid with 10 end-of-month payments. The monthly interest rate is $2\%$ . If each payment is $P$ dollars, the value of all payments at the time of the last payment is $P + P(1.02) + P(1.02)^2 + \cdots + P(1.02)^9.$ Which expression correctly represents this sum in closed form?
Answer: $P\,\dfrac{1-(1.02)^{10}}{1-1.02}$ 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. In loan calculations, payments grow by the interest factor when valued at a future time: the series $P + P(1.02) + P(1.02)^2 + \cdots + P(1.02)^9$ represents 10 payments valued at the time of the last payment. To identify parameters: a = P (the last payment, made "now" with no growth), r = 1.02 (each earlier payment has grown by 2% per month), and n = 10 (we have 10 total payments). The 10th payment (just made) = $P(1.02)^0 = P$ , the 9th payment = $P(1.02)^1$ , ..., the 1st payment = $P(1.02)^9$ . Applying the formula: $S_{10} = P \frac{1 - (1.02)^{10}}{1 - 1.02}$ . Choice B correctly shows this expression with n = 10. Choice A incorrectly uses n = 9, forgetting that 10 payments means n = 10 in the formula, even though the highest power shown is 9. This is the classic "off by one" error in financial series! The three-step strategy for loan problems: (1) Count total payments (here 10). (2) Identify which payment is valued "now" (usually power 0). (3) Apply formula with n = total number of payments. Financial geometric series often have highest power = (n-1) because we start counting at power 0!

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