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Algebra Help: Deriving Applying The Geometric Series Formula

Review real example questions for Deriving Applying The Geometric Series Formula in Algebra.

Question 1 / 10

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

All questions

Question 1

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

Explanation: This question tests your understanding of geometric series—the sum of terms from a geometric sequence—and how to use the formula $S_n = a(1 - r^n)/(1 - r)$ 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 $S_n = a + ar + ar^2 + \cdots + ar^{n-1}$ , then multiply by $r$ to get $rS_n = ar + ar^2 + ar^3 + \cdots + ar^n$ . 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: $S_n = a + ar + ar^2 + \cdots + ar^{n-1}$ . (2) Multiply both sides by $r$ : $rS_n = ar + ar^2 + ar^3 + \cdots + ar^n$ . (3) Subtract second from first: $S_n - rS_n = (a + ar + ar^2 + \cdots + ar^{n-1}) - (ar + ar^2 + ar^3 + \cdots + ar^n)$ . (4) Notice the cancellation: all middle terms cancel, leaving $S_n - rS_n = a - ar^n$ . (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 the accurate final expression. Choice A makes an error in the derivation by using $r^{n-1}$ instead of $r^n$ in the numerator; this happens if you miss that the remaining term after cancellation is $-ar^n$ from the shifted series. 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 'ar^2 $and$ rS_n $has 'ar^2' (cancel!), etc. Only a from$ S_n $and$ -ar^n $from$ rS_n$ don't cancel. This telescoping is the insight! Once you see it, you'll never forget the derivation.

Question 2

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 $a=10$ , $r=0.8$ , $n=5$ ).

$\dfrac{10(1-0.8^5)}{1-0.8}=33.616$ (correct answer)
$\dfrac{10(1-0.8^4)}{1-0.8}=29.52$
$\dfrac{10(1-0.8^5)}{1-0.8}=16.808$
$10(0.8)^5=3.2768$

Explanation: This question tests your understanding of geometric series—the sum of terms from a geometric sequence—and how to use the formula S_n = a(1 - $r^n$ )/(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!

Question 3

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 $a=10$ , $r=0.8$ , $n=5$ .

$\approx 33.62$ (correct answer)
$\approx 40.00$
$\approx 26.89$
$\approx 36.16$

Explanation: This question tests your understanding of geometric series—the sum of terms from a geometric sequence—and how to use the formula S_n = a(1 - $r^n$ )/(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.

Question 4

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).

$\$ 1{,}200.00$
$\$ 1{,}236.94$ (correct answer)
$\$ 1{,}260.00$
$\$ 1{,}230.79$

Explanation: This question tests your understanding of geometric series—the sum of terms from a geometric sequence—and how to use the formula S_n = a(1 - $r^n$ )/(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.

Question 5

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 $a=7$ , $r=-0.5$ , $n=8$ .)

$\dfrac{120}{16}$
$\dfrac{1799}{128}$
$\dfrac{119}{16}$ (correct answer)
$\dfrac{63}{8}$

Explanation: This question tests your understanding of geometric series—the sum of terms from a geometric sequence—and how to use the formula S_n = a(1 - $r^n$ )/(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).

Question 6

For the geometric series $2 + 6 + 18 + 54$ , identify $a$ , $r$ , and $n$ , then find the sum using $S_n = \frac{a(1-r^n)}{1-r}$ .

$a=2, r=2, n=4; S_4=30$
$a=2, r=3, n=4; S_4=80$ (correct answer)
$a=2, r=3, n=3; S_3=26$
$a=6, r=3, n=4; S_4=240$

Explanation: This question tests your understanding of geometric series—the sum of terms from a geometric sequence—and how to use the formula $S_n = \frac{a(1 - r^n)}{1 - r}$ to calculate these sums efficiently. A geometric series is the sum of terms from a geometric sequence: if the sequence is 2, 6, 18, 54, ... (multiply by 3 each time), the series is 2 + 6 + 18 + 54 + ... (adding those terms up). Rather than adding manually (tedious for many terms!), we use the formula $S_n = \frac{a(1 - r^n)}{1 - r}$ , where a is the first term, r is the common ratio, and n is how many terms we're summing. This formula works for any finite geometric series! Let's verify the formula works for a simple example: series 2 + 6 + 18 + 54 with a = 2, r = 3, n = 4. Formula: $S_4 = 2(1 - 3^4)/(1 - 3) = 2(1 - 81)/(-2) = 2(-80)/(-2) = 80$ . 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 A correctly identifies a=2, r=3, n=4 and calculates the sum as 80 with accurate arithmetic. Choice B identifies the parameters wrong: n=3 instead of n=4; getting a, r, or n wrong throws off the entire calculation! Why this formula is powerful: to sum 2 + 6 + 18 + 54 + ... + ( $2·3^99$ ), 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!

Question 7

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

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

$484$ (correct answer)
$480$
$364$
$972$

Explanation: This question tests your understanding of geometric series—the sum of terms from a geometric sequence—and how to use the formula S_n = a(1 - $r^n$ )/(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 4 + 12 + 36 + 108 + 324 with a = 4, r = 3, n = 5: Using the formula S_n = a(1 - $r^n$ )/(1 - r), we substitute: S_5 = 4(1 - $3^5$ )/(1 - 3) = 4(1 - 243)/(-2) = 4(-242)/(-2) = 4(121) = 484. So the sum of the first 5 terms is 484. Choice A correctly applies the formula with a = 4, r = 3, n = 5 and calculates the sum as 484 with accurate arithmetic. Choice C gives 364, which might come from a calculation error with 3^5: if someone calculated 3^5 as 183 instead of 243, they'd get 4(1-183)/(-2) = 4(-182)/(-2) = 364. With geometric series, r^n can get large quickly ( $2^10$ = 1024!), so calculator use is often needed. Double-check power calculations! Let's verify the formula works for a simple example: series 2 + 6 + 18 + 54 with a = 2, r = 3, n = 4. Formula: S_4 = 2(1 - $3^4$ )/(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!

Question 8

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 $S_n=\frac{a(1-r^n)}{1-r}$ .

$80$ (correct answer)
$78$
$72$
$162$

Explanation: This question tests your understanding of geometric series—the sum of terms from a geometric sequence—and how to use the formula $S_n = a(1 - r^n)/(1 - r)$ 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 $r^{n-1}$ .

Question 9

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 $a=3$ , $r=2$ , $n=5$ .

$90$
$93$ (correct answer)
$96$
$189$

Explanation: This question tests your understanding of geometric series—the sum of terms from a geometric sequence—and how to use the formula $S_n = a(1 - r^n)/(1 - r)$ 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!

Question 10

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

$315$ (correct answer)
$320$
$310$
$63$

Explanation: This question tests your understanding of geometric series—the sum of terms from a geometric sequence—and how to use the formula S_n = a(1 - $r^n$ )/(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!