Deriving/Applying the Geometric Series Formula - Algebra 1

Use the geometric series formula $S_n=\dfrac{a(1-r^n)}{1-r}$ to calculate the sum $3+3(1.1)+3(1.1)^2+\cdots+3(1.1)^9.$ (So $a=3$ , $r=1.1$ , $n=10$ .)
Answer: $\dfrac{3\left(1-1.1^{10}\right)}{1-1.1}$ 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 - $r^n$ )/(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 $S_n$ = 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 $S_5$ = 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 $S_n$ = a(1 - $r^n$ )/(1 - r), we substitute: $S_{10}$ = 3(1 - $1.1^10$ )/(1 - 1.1). This matches the expression in choice A exactly. So the sum is expressed as 3(1 - $1.1^10$ )/(1 - 1.1). Choice A correctly applies the formula with a = 3, r = 1.1, n = 10 with accurate substitution. Choice B has n = 9 instead of n = 10: when counting from 3(1.1)⁰ to 3(1.1)⁹, that's 10 terms total (including the 0th power). Remember that powers go from 0 to n-1, giving n terms total. Off-by-one errors in counting terms are common! 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 $r^{n-1}$ .
Verify a computation using the geometric series formula: Evaluate $S_5=3+6+12+24+48$ by using $S_n=\frac{a(1-r^n)}{1-r}$ with $a=3$ , $r=2$ , $n=5$ .
Answer: $93$ 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 - $r^n$ )/(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: $S_4$ = 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 $S_5$ = 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 - $r^n$ ) 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 - $r^n$ )/(1 - r) and calculate carefully, watching signs: when r > 1, both (1 - $r^n$ ) and (1 - r) are negative; when 0 < r < 1, both are positive. The formula handles all cases!
A ball's rebound heights form a geometric sequence: after the first bounce it rises to $10$ ft, and each bounce reaches $80\%$ of the previous height. What is the sum of the heights of the first 5 bounces (in feet)? This is the geometric series $10+10(0.8)+10(0.8)^2+10(0.8)^3+10(0.8)^4$ with $a=10$ , $r=0.8$ , $n=5$ .
Answer: $\approx 33.62$ 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 - $r^n$ )/(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! 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 - $0.8^5$ )/(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 - $r^n$ )/(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.
Calculate the sum of the geometric series $2+6+18+54$ . (This is a geometric series, meaning a sum; the corresponding sequence is $2,6,18,54$ .) Use $a=2$ , $r=3$ , $n=4$ and $S_n=\frac{a(1-r^n)}{1-r}$ .
Answer: $80$ This question tests your understanding of geometric series—the sum of terms from a geometric sequence—and how to use the formula $S_n = a(1 - r^n)/(1 - r)$ to calculate these sums efficiently. A geometric series is the sum of terms from a geometric sequence: if the sequence is $2, 6, 18, 54, \dots$ (multiply by 3 each time), the series is $2 + 6 + 18 + 54 + \dots$ (adding those terms up). Rather than adding manually (tedious for many terms!), we use the formula $S_n = a(1 - r^n)/(1 - r)$ , 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 $2 + 6 + 18 + 54$ with a = 2, r = 3, n = 4. Formula: $S_4 = 2(1 - 3^4)/(1 - 3) = 2(1 - 81)/(-2) = 2(-80)/(-2) = 80$ . 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 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: $6/2 = 3$ , 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 $a + ar + ar^2 + \dots + ar^{n-1}$ , 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 $ar^4$ , not $ar^5$ . Off-by-one errors in n throw off the sum. The formula already accounts for this with $r^n$ , not $r^{n-1}$ .
Verify a computation using the geometric series formula: Evaluate $S_5=3+6+12+24+48$ by using $S_n=\frac{a(1-r^n)}{1-r}$ with $a=3$ , $r=2$ , $n=5$ .
Answer: $93$ This question tests your understanding of geometric series—the sum of terms from a geometric sequence—and how to use the formula $S_n = a(1 - r^n)/(1 - r)$ 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: $S_4 = 2(1 - 3^4)/(1 - 3) = 2(1 - 81)/(-2) = 2(-80)/(-2) = 80$ . 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 $S_5 = 3(1 - 2^5)/(1 - 2) = 3(1 - 32)/(-1) = 3(-31)/(-1) = 3(31) = 93$ . 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 - $r^n$ ) 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 $S_n = a(1 - r^n)/(1 - r)$ and calculate carefully, watching signs: when r > 1, both (1 - $r^n$ ) and (1 - r) are negative; when 0 < r < 1, both are positive. The formula handles all cases!
Use the geometric series sum formula $S_n=\frac{a(1-r^n)}{1-r}$ (for $r\ne 1$ ) to find the sum of the first 6 terms of the geometric series $5+10+20+40+\cdots$ . (Here $a=5$ , $r=2$ , $n=6$ .)
Answer: $315$ 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 - $r^n$ )/(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 $S_n$ = a(1 - $r^n$ )/(1 - r), (5) calculate carefully. For the series 5 + 10 + 20 + 40 + ... with a = 5, r = 2, n = 6: Using the formula $S_6$ = 5(1 - $2^6$ )/(1 - 2), we substitute: $S_6$ = 5(1 - 64)/(-1) = 5(-63)/(-1) = 5(63) = 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 has a sign error: when r > 1, we have negative numerator and negative denominator: (1 - $r^n$ ) is negative (1 - big number), and (1 - r) is negative (1 - bigger than 1), so negative/negative = positive, but this choice likely forgot the signs and got half the value or similar. 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 - $r^n$ )/(1 - r) and calculate carefully, watching signs: when r > 1, both (1 - $r^n$ ) and (1 - r) are negative; when 0 < r < 1, both are positive. The formula handles all cases!
A bouncing ball rises to $80\%$ of its previous height after each bounce. If the first bounce reaches $10$ ft, what is the total of the bounce heights for the first $5$ bounces? Sum the geometric series $10+10(0.8)+10(0.8)^2+10(0.8)^3+10(0.8)^4$ (so $a=10$ , $r=0.8$ , $n=5$ ).
Answer: $\dfrac{10(1-0.8^5)}{1-0.8}=33.616$ 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 - $r^n$ )/(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! 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 - $r^n$ ) 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 + ... + ( $2·3^99$ ), you'd need to add 100 terms manually (impossible!). With the formula: $S_{100}$ = 2(1 - $3^100$ )/(1 - 3) and you're done (calculator handles $3^100$ ). The formula turns a hundred-operation problem into a few operations. That's the beauty of having a formula!
Apply the geometric series formula to a savings plan: You deposit $100 at the end of each month into an account earning $0.5\%$ interest per month. What is the total value of the 12 deposits immediately after the 12th deposit is made? Model as a finite geometric series with $a=100$ , $r=1.005$ , $n=12$ : $S_{12}=100+100(1.005)+\cdots+100(1.005)^{11}.$
Answer: $100\cdot\dfrac{1-(1.005)^{12}}{1-1.005}\approx 1233.01$ 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 - $r^n$ )/(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! In the context of a savings plan, each payment of $100 grows by 0.5%, creating a geometric series: the first deposit earns interest for 11 months (100 $(1.005)^{11}$ ), the second for 10 months (100 $(1.005)^{10}$ ), down to the last deposit earning no interest (100). With a = 100, r = 1.005, n = 12, the total is $S_{12}$ = 100(1 - $(1.005)^{12}$ )/(1 - 1.005) = 100(1 - 1.0616...)/(−0.005) = 100(−0.0616...)/(−0.005) = 100(12.3301...) ≈ 1233.01. This shows the account value after all deposits. Choice A correctly applies the formula with a = 100, r = 1.005, n = 12 and calculates the sum as approximately 1233.01 with accurate arithmetic. Choice C has the denominator inverted: it shows (1.005 - 1) instead of (1 - 1.005), which would give a positive denominator when it should be negative. With r > 1, we need (1 - r) in the denominator, which gives us a negative value. The negative numerator divided by negative denominator gives the correct positive sum. 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.
Use $S_n=\frac{a(1-r^n)}{1-r}$ to find the sum: $10+5+2.5+1.25+0.625+0.3125.$ (So $a=10$ , $r=\tfrac12$ , $n=6$ .)
Answer: $19.6875$ 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 - $r^n$ )/(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 $S_n$ = 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 $S_5$ = 5(1 - $2^5$ )/(1 - 2) = 5(1 - 32)/(-1) = 5(-31)/(-1) = 155. For the series 10 + 5 + 2.5 + 1.25 + 0.625 + 0.3125 with a = 10, r = 1/2, n = 6: Using the formula $S_n$ = a(1 - $r^n$ )/(1 - r), we substitute: $S_6$ = 10(1 - $(1/2)^6$ )/(1 - 1/2) = 10(1 - 1/64)/(1/2) = 10(63/64)/(1/2) = 10(63/64)(2) = 10(126/64) = 1260/64 = 19.6875. So the sum of the first 6 terms is 19.6875. Choice A correctly applies the formula with a = 10, r = 1/2, n = 6 and calculates the sum as 19.6875 with accurate arithmetic. Choice B gives 20, which might be tempting as a round number, but the actual sum is 19.6875. When r < 1, the series converges toward a limit, and with 6 terms we're close but not quite at 20. Don't round prematurely in exact 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 - $r^n$ )/(1 - r) and calculate carefully, watching signs: when r > 1, both (1 - $r^n$ ) and (1 - r) are negative; when 0 < r < 1, both are positive. The formula handles all cases!
Find the sum of the first 6 terms of the geometric series $5+10+20+40+80+160.$ (Here, $a=5$ , $r=2$ , $n=6$ .)
Answer: $315$ 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 - $r^n$ )/(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 $S_n$ = 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! For the series 5 + 10 + 20 + 40 + 80 + 160 with a = 5, r = 2, n = 6: Using the formula $S_n$ = a(1 - $r^n$ )/(1 - r), we substitute: $S_6$ = 5(1 - $2^6$ )/(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 - $2^7$ )/(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 - $r^n$ )/(1 - r) and calculate carefully, watching signs: when r > 1, both (1 - $r^n$ ) and (1 - r) are negative; when 0 < r < 1, both are positive. The formula handles all cases!
A ball's rebound heights form a geometric sequence: after the first bounce it rises to $10$ ft, and each bounce reaches $80\%$ of the previous height. What is the sum of the heights of the first 5 bounces (in feet)? This is the geometric series $10+10(0.8)+10(0.8)^2+10(0.8)^3+10(0.8)^4$ with $a=10$ , $r=0.8$ , $n=5$ .
Answer: $\approx 33.62$ 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 - $r^n$ )/(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! 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 - $0.8^5$ )/(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 - $r^n$ )/(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.
Derive the finite geometric series sum formula. Start with $S_n=a+ar+ar^2+\cdots+ar^{n-1}\quad (r\ne 1)$ Multiply both sides by $r$ and subtract to solve for $S_n$ . Which expression for $S_n$ 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 use the formula $S_n$ = a(1 - $r^n$ )/(1 - r) to calculate these sums efficiently. The geometric series formula $S_n$ = a(1 - $r^n$ )/(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 - $r^n$ ), so $S_n$ = a(1 - $r^n$ )/(1 - r). The middle terms canceling is the magic that makes this work! Deriving $S_n$ = a(1 - $r^n$ )/(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 + ... + $ar^n$ ). (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 - $r^n$ ). (6) Divide by (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 arithmetic. Choice A has an error in the exponent: when we have n terms from a to $ar^{n-1}$ , the subtraction gives us a - $ar^n$ , not a - $ar^{n-1}$ , so the formula has (1 - $r^n$ ), not (1 - $r^{n-1}$ ). Don't confuse the number of terms with the last exponent! 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.
A ball rebounds to $80\%$ of its previous height. The first rebound reaches 10 ft, the second reaches $10(0.8)$ ft, and so on. Find the sum of the rebound heights for the first 6 rebounds: $10+10(0.8)+10(0.8)^2+\cdots+10(0.8)^5.$
Answer: $\dfrac{10(1-0.8^6)}{1-0.8}=50(1-0.8^6)$ 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 - $r^n$ )/(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 $S_n$ = 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! For the series 10 + 10(0.8) + 10(0.8)² + ... + 10(0.8)⁵ with a = 10, r = 0.8, n = 6: Using the formula $S_n$ = a(1 - $r^n$ )/(1 - r), we substitute: $S_6$ = 10(1 - $0.8^6$ )/(1 - 0.8) = 10(1 - $0.8^6$ )/0.2. Since 1/0.2 = 5, this equals 10 × 5 × (1 - $0.8^6$ ) = 50(1 - $0.8^6$ ). So the sum of the first 6 rebounds is 50(1 - $0.8^6$ ). Choice A correctly applies the formula with a = 10, r = 0.8, n = 6 and simplifies to show the sum as 50(1 - $0.8^6$ ) with accurate arithmetic. Choice B uses n = 5 instead of n = 6: the problem asks for 6 rebounds, from the first rebound (10 ft) through the sixth rebound (10 $(0.8)^5$ ft). That's 6 terms total, so n = 6. Count the rebounds carefully—don't miss the last one! Why this formula is powerful: to sum 2 + 6 + 18 + 54 + ... + ( $2·3^99$ ), you'd need to add 100 terms manually (impossible!). With the formula: $S_{100}$ = 2(1 - $3^100$ )/(1 - 3) and you're done (calculator handles $3^100$ ). The formula turns a hundred-operation problem into a few operations. That's the beauty of having a formula!
Use $S_n=\dfrac{a(1-r^n)}{1-r}$ to find the sum of the geometric series $7+7(-0.5)+7(-0.5)^2+\cdots+7(-0.5)^7.$ (So $a=7$ , $r=-0.5$ , $n=8$ .)
Answer: $\dfrac{119}{16}$ 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 - $r^n$ )/(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 $S_n$ = 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! 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 - $r^n$ )/(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 $r^{n-1}$ .
In a savings plan, you deposit $100 at the end of each month. The account earns $0.5\%$ interest per month. After 12 months, the value of the deposits is $100\big(1.005^{11}+1.005^{10}+\cdots+1.005^1+1.005^0\big).$ Use $S_n=\dfrac{a(1-r^n)}{1-r}$ to find the total value after 12 months (in dollars, to the nearest cent).
Answer: $\$1{,}236.94$ 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 - $r^n$ )/(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! 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 - $1.005^12$ )/(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( $1.005^0$ + $1.005^1$ + ... + $1.005^11$ ). Using a = 100, r = 1.005, n = 12: $S_{12}$ = 100( $1.005^12$ - 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.
Derive the finite geometric series sum formula. Starting from $S_n=a+ar+ar^2+\cdots+ar^{n-1},$ which expression correctly results from multiplying by $r$ and subtracting to eliminate the middle terms (assume $r\ne 1$ )?
Answer: $S_n(1-r)=a(1-r^n)$ , so $S_n=\dfrac{a(1-r^n)}{1-r}$ This question tests your understanding of geometric series—the sum of terms from a geometric sequence—and how to use the formula $S_n$ = a(1 - $r^n$ )/(1 - r) to calculate these sums efficiently. The geometric series formula $S_n$ = a(1 - $r^n$ )/(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 - $r^n$ ), so $S_n$ = a(1 - $r^n$ )/(1 - r). The middle terms canceling is the magic that makes this work! Deriving $S_n$ = a(1 - $r^n$ )/(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 + ... + $ar^n$ ). (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 - $r^n$ ). (6) Divide by (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 $S_n$ (1 - r) = a(1 - $r^n$ ), so $S_n$ = a(1 - $r^n$ )/(1 - r) with accurate arithmetic. Choice D makes an error in the derivation: when subtracting $S_n$ - rS_n, the last term of $S_n$ is $ar^{n-1}$ and the last term of rS_n is $ar^n$ , so we get a - $ar^n$ , not a - $ar^{n-1}$ . The key insight is that ALL terms except the first and last cancel! 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.
Use the geometric series sum formula $S_n=\frac{a(1-r^n)}{1-r}$ (for $r\ne 1$ ) to calculate $3 + 3(1.1) + 3(1.1)^2 + \cdots + 3(1.1)^9.$ (So $a=3$ , $r=1.1$ , $n=10$ .)
Answer: $\dfrac{3\big(1-(1.1)^{10}\big)}{1-1.1}\approx 47.81$ 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 - $r^n$ )/(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 $S_n$ = 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! 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 $S_n$ = a(1 - $r^n$ )/(1 - r), we substitute: $S_{10}$ = 3(1 - $(1.1)^{10}$ )/(1 - 1.1) = 3(1 - 2.5937...)/(−0.1) = 3(−1.5937...)/(−0.1) = 3(15.937...) ≈ 47.81. So the sum of the first 10 terms is approximately 47.81. Choice A correctly applies the formula with a = 3, r = 1.1, n = 10 and calculates the sum as approximately 47.81 with accurate arithmetic. Choice C has the denominator inverted: it shows (1.1 - 1) instead of (1 - 1.1), which would give a positive denominator when it should be negative. With r > 1, we need (1 - r) in the denominator, which gives us a negative value. The negative numerator divided by negative denominator gives the correct positive sum. Why this formula is powerful: to sum 2 + 6 + 18 + 54 + ... + ( $2·3^99$ ), you'd need to add 100 terms manually (impossible!). With the formula: $S_{100}$ = 2(1 - $3^100$ )/(1 - 3) and you're done (calculator handles $3^100$ ). The formula turns a hundred-operation problem into a few operations. That's the beauty of having a formula!
What is the sum of the first 8 terms of the geometric series $7+14+28+\cdots$ ? (So $a=7$ , $r=2$ , $n=8$ .)
Answer: $1785$ 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 - $r^n$ )/(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 $S_n$ = a(1 - $r^n$ )/(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 - $2^8$ )/(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 - $r^n$ ) 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 + ... + ( $2·3^99$ ), you'd need to add 100 terms manually (impossible!). With the formula: $S_{100}$ = 2(1 - $3^100$ )/(1 - 3) and you're done (calculator handles $3^100$ ). The formula turns a hundred-operation problem into a few operations. That's the beauty of having a formula!
What is the sum of the first 5 terms of the geometric series $2+6+18+54+162?$ (Identify $a=2$ , $r=3$ , $n=5$ .)
Answer: $242$ 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 - $r^n$ )/(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 $S_n$ = 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! For the series 2 + 6 + 18 + 54 + 162 with a = 2, r = 3, n = 5: Using the formula $S_n$ = a(1 - $r^n$ )/(1 - r), we substitute: $S_5$ = 2(1 - $3^5$ )/(1 - 3) = 2(1 - 243)/(-2) = 2(-242)/(-2) = 2(121) = 242. So the sum of the first 5 terms is 242. Choice A correctly applies the formula with a = 2, r = 3, n = 5 and calculates the sum as 242 with accurate arithmetic. Choice C gives 121, which is exactly half of the correct answer 242. This suggests forgetting to multiply by a = 2 in the final step: (1 - $3^5$ )/(1 - 3) = (-242)/(-2) = 121, but we need 2 × 121 = 242. Don't forget the first term multiplier a in the formula! 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 - $3^4$ )/(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!
Use the geometric series sum formula $S_n = \frac{a(1-r^n)}{1-r}$ (for $r \ne 1$ ) to find the sum of the first $6$ terms of the geometric series $5 + 10 + 20 + 40 + 80 + 160.$ Clearly, $a=5$ , $r=2$ , and $n=6$ .
Answer: $315$ This question tests your understanding of geometric series—the sum of terms from a geometric sequence—and how to use the formula $S_n = a(1 - r^n)/(1 - r)$ 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 $S_n = a(1 - r^n)/(1 - r)$ , (5) calculate carefully. Example: for 5 + 10 + 20 + 40 + 80, we have a = 5, r = 2, n = 5, so $S_5 = 5(1 - 2^5)/(1 - 2) = 5(1 - 32)/(-1) = 5(-31)/(-1) = 155$ . For the series 5 + 10 + 20 + 40 + 80 + 160 with a = 5, r = 2, n = 6: Using the formula $S_n = a(1 - r^n)/(1 - r)$ , we substitute: $S_6 = 5(1 - 2^6)/(1 - 2) = 5(1 - 64)/(-1) = 5(-63)/(-1) = 315$ . 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 $r \neq 1$ (if r = 1, sum = na instead). (3) Substitute into $S_n = a(1 - r^n)/(1 - r)$ and calculate carefully, watching signs: when r > 1, both (1 - $r^n$ ) and (1 - r) are negative; when 0 < r < 1, both are positive. The formula handles all cases!

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