C. Discrete Functions · Sequences and SeriesSolving Problems with Arithmetic and Geometric SeriesC2.4 C2.3About 75 minutesLearning goalsI can recognize when a real problem asks for a sum (a series) rather than a single term.I can model real situations with arithmetic or geometric series and solve them.I can find the total distance travelled by a bouncing ball.I can find how many terms are needed to reach a total.Words to knowtotalA word that usually signals a series: you add up terms instead of finding one term.cumulativeAdding up as you go, such as the total distance run so far. Cumulative totals are partial sums Sₙ.One term or the total?In the last lesson you answered questions like "How high is the 5th bounce?" or "How many seats are in row 18?" Those ask for one term. Many real questions ask for a total instead: How far does the ball travel altogether? How many seats are in the whole section? How many people have been called by the end of the phone tree? Those are series questions.The plan is the same as before, with one extra decision:Arithmetic or geometric? Find a and d or r.One term (tₙ) or a total (Sₙ)? Words like total, altogether, in all, so far point to a sum.Find n carefully.Use the right formula and answer in a sentence with units.Worked exampleExample 1: A pile of logsLogs for a cottage woodstove are piled with 20 logs in the bottom row, 19 in the next, and so on, with 5 logs in the top row. How many logs are in the pile?Each row has 1 fewer log: arithmetic, a = 20, d = −1, last term 5. We want the total: a series.Number of rows: 5 = 20 + (n − 1)(−1) → n − 1 = 15 → n = 16.S₁₆ = 16(20 + 5) ÷ 2 = 8(25) = 200 logs.200 logs8 logs7 logs6 logs5 logs4 logsRows: 8, 7, 6, 5, 4an arithmetic serieswith n = 5, a = 8, t₅ = 4S₅ = 5(8 + 4) ÷ 2 = 30 logsA smaller pile with rows of 8, 7, 6, 5 and 4 logs. Adding the rows is an arithmetic series: 5 rows with an average of (8 + 4) ÷ 2 = 6 logs, so 30 logs.Worked exampleExample 2: A falling objectIgnoring air resistance, an object dropped from rest falls about 4.9 m in the 1st second, 14.7 m in the 2nd second, 24.5 m in the 3rd second, and so on. How far does it fall in the first 8 seconds?Differences: 14.7 − 4.9 = 9.8 and 24.5 − 14.7 = 9.8. Arithmetic, a = 4.9, d = 9.8, n = 8.S₈ = 8[2(4.9) + 7(9.8)] ÷ 2 = 4[9.8 + 68.6] = 4(78.4) = 313.6 m.(Physics gives the same answer as 4.9t² = 4.9(64) = 313.6 m. Real objects fall less far because of air resistance.)313.6 mWorked exampleExample 3: Total distance of a bouncing ballA ball is dropped from 200 cm and rebounds to 75% of its previous height each time. What total vertical distance has it travelled when it hits the ground for the 6th time?Draw it! The ball first falls 200 cm (1st hit). After that, each bounce goes up and then the same distance down. The 6th hit comes at the end of the 5th bounce.Rebound heights: 150, 112.5, 84.375, … a geometric sequence with a = 150, r = 0.75. Sum of the first 5: S₅ = 150(1 − 0.75⁵) ÷ (1 − 0.75) = 600(1 − 0.2373…) ≈ 457.62 cm.Each rebound height is travelled twice (up and down): 2 × 457.62 = 915.23 cm.Total = 200 + 915.23 ≈ 1115.2 cm, about 11.2 m.about 1115 cm (11.2 m)Worked exampleExample 4: A phone treeTo cancel practice on a snow day, a coach calls 3 players. In each round, every player who was just called phones 3 new players. (a) How many players have been called after 4 rounds? (b) How many rounds are needed to reach all 1000 students at the school?Calls per round: 3, 9, 27, 81, …: geometric, a = 3, r = 3. Asking for everyone called so far means a sum.(a) S₄ = 3(3⁴ − 1) ÷ 2 = 3(80) ÷ 2 = 120 players.(b) Need Sₙ = 3(3ⁿ − 1) ÷ 2 ≥ 1000 → 3ⁿ − 1 ≥ 666.7 → 3ⁿ ≥ 667.7. 3⁵ = 243 (too small), 3⁶ = 729 (enough). So 6 rounds (S₆ = 3(728) ÷ 2 = 1092 calls; in reality some calls would go unanswered or to people already reached).(a) 120 (b) 6 roundscoachround 1: 3round 2: 9round 3: 27Total calls after 3 rounds: 3 + 9 + 27 = 39The first three rounds of the phone tree: 3, then 9, then 27 players are called. The total so far is a geometric series.Worked exampleExample 5: A training planNoor runs 5 km in her first week of training and increases her weekly distance by 10% each week. What total distance does she run in the first 10 weeks?Weekly distances: 5, 5.5, 6.05, …: geometric, a = 5, r = 1.1. Total over 10 weeks: a sum.S₁₀ = 5(1.1¹⁰ − 1) ÷ (1.1 − 1) = 5(2.5937… − 1) ÷ 0.1 = 50(1.5937…) ≈ 79.7 km.about 79.7 kmWorked exampleExample 6: Reaching a savings goalLiam saves $5 in the first week, $7 the next, and $2 more each week after that, in a jar (no interest). After how many weeks will he have saved at least $500?Arithmetic, a = 5, d = 2. Total saved after n weeks: Sₙ = n[10 + 2(n − 1)] ÷ 2 = n(n + 4) = n² + 4n.Solve n² + 4n ≥ 500. Try n = 20: 400 + 80 = 480 (not yet). n = 21: 441 + 84 = 525 ✓.After 21 weeks he has $525. (Solving n² + 4n − 500 = 0 with the quadratic formula gives n ≈ 20.4, so 21 full weeks.)21 weeksComing up in Financial ApplicationsWhen you deposit the same amount regularly into an account that earns compound interest, each deposit grows by a different number of compounding periods. Adding them up gives a geometric series. That is exactly how annuities work, and you will use the geometric series formula again in the unit Financial Applications.Just for funIn math problems, a ball can bounce forever, losing a quarter of its height each time.In real life, it bounces about a dozen times and then rolls under the couch, where it stays until spring cleaning.Hands-on activityBuild a cup pyramidYou need: 10 to 21 identical paper or plastic cups and a flat floor or table.Build a pyramid with 4 cups on the bottom row, 3 on the next, then 2, then 1. Count the cups. Check with Sₙ = n(a + tₙ) ÷ 2.Predict how many cups you need for a 6-row pyramid. Build it if you have enough cups.A cup-stacking team wants a pyramid with 15 cups on the bottom row. How many cups do they need?Reverse the question: you have 50 cups. What is the tallest complete pyramid you can build?What to look for: 4 + 3 + 2 + 1 = 10 cups = 4(4 + 1) ÷ 2 ✓. 6 rows: 6(7) ÷ 2 = 21 cups. Bottom row 15: 15(16) ÷ 2 = 120 cups. With 50 cups: n(n + 1) ÷ 2 ≤ 50 → 9 rows need 45 and 10 rows need 55, so 9 rows (with 5 cups left over).Your turn1A section of an arena has 25 rows. Row 1 has 16 seats and each row has 2 more than the one before. How many seats are in the section?Show answerS₂₅ = 25[32 + 24(2)] ÷ 2 = 25(80) ÷ 2 = 1000 seats.2A stack of pipes has 15 on the bottom and 1 fewer in each row, up to 1 on top. How many pipes are there?Show answer15(15 + 1) ÷ 2 = 120 pipes.3A ball is dropped from 4 m and rebounds to 50% of its height each time. What total distance has it travelled when it hits the ground for the 4th time?Show answerDown 4 m, then bounces of 2, 1, 0.5 m up and down: 4 + 2(2 + 1 + 0.5) = 4 + 7 = 11 m.4A chain message is sent to 2 people, each sends it to 2 more, and so on. How many messages are sent in the first 10 rounds?Show answerS₁₀ = 2(2¹⁰ − 1) = 2(1023) = 2046 messages.5A cyclist rides 20 km on day 1 and 3 km more each day. What is her total distance over 14 days?Show answerS₁₄ = 14[40 + 13(3)] ÷ 2 = 7(79) = 553 km.6A company's sales were $50 000 in its first year and grew by 8% each year. What were its total sales over the first 6 years?Show answerS₆ = 50 000(1.08⁶ − 1) ÷ 0.08 ≈ 50 000(0.58687) ÷ 0.08 ≈ $366 796.More practice: warm-up1"How many seats are in row 12?" Term or sum?Show answerTerm (t₁₂)2"How many seats are in rows 1 to 12?" Term or sum?Show answerSum (S₁₂)3Find 3 + 6 + 9 + … + 30.Show answer10 terms: 10(3 + 30) ÷ 2 = 1654Find 1 + 2 + 4 + … + 512.Show answer10 terms: 2¹⁰ − 1 = 10235A ball is dropped from 1 m and hits the ground once. How far has it travelled?Show answer1 m (no bounces yet).More practice: on your own1A display of soup cans has 1 can on top, 3 in the next row, 5 in the next, and so on, with 12 rows. How many cans are in the display?Show answerArithmetic, a = 1, d = 2, n = 12: S₁₂ = 12[2 + 11(2)] ÷ 2 = 6(24) = 144 cans (= 12², the sum of the first 12 odd numbers).2A new bakery sells 40 muffins on its first day and 6 more each day after. On which day does the total number sold first pass 2000?Show answerSₙ = n[80 + 6(n − 1)] ÷ 2 = n(3n + 37) = 3n² + 37n. n = 20: 1200 + 740 = 1940. n = 21: 1323 + 777 = 2100. Day 21.3A ball is dropped from 10 m and rebounds to 80% of its previous height. How far has it travelled when it hits the ground for the 5th time?Show answer10 + 2(8 + 6.4 + 5.12 + 4.096) = 10 + 2(23.616) = 57.232 m, about 57.2 m. (Formula: S₄ = 8(1 − 0.8⁴) ÷ 0.2 = 40(0.5904) = 23.616.)4A trail-building crew clears 600 m of trail in its first week and 90% as much each week after that. How much trail is cleared in total in 8 weeks?Show answerS₈ = 600(1 − 0.9⁸) ÷ 0.1 = 6000(1 − 0.43047) ≈ 3417 m (about 3.4 km).5How many terms of the geometric series 4 + 12 + 36 + … are needed for the sum to exceed 10 000?Show answerSₙ = 4(3ⁿ − 1) ÷ 2 = 2(3ⁿ − 1) > 10 000 → 3ⁿ > 5001. 3⁷ = 2187, 3⁸ = 6561. 8 terms (S₈ = 13 120).Challenge1A ball dropped from 200 cm rebounds to 75% of its height forever. Explain why the total distance approaches 1400 cm.Show answerThe rebound heights sum to Sₙ = 150(1 − 0.75ⁿ) ÷ 0.25 = 600(1 − 0.75ⁿ). As n grows, 0.75ⁿ → 0, so the rebound total approaches 600 cm. Each rebound is travelled up and down: 2 × 600 = 1200, plus the first 200 cm drop: 1400 cm.2An auditorium has 20 rows. The front row has x seats and each row has 3 more than the row before. The auditorium has 1070 seats. How many seats are in the front row and in the back row?Show answerS₂₀ = 20[2x + 19(3)] ÷ 2 = 10(2x + 57) = 20x + 570 = 1070 → 20x = 500 → x = 25 seats in the front row; back row t₂₀ = 25 + 19(3) = 82 seats.3In a tournament bracket, 64 teams play in round 1, and half the teams are eliminated each round until one champion remains. How many games are played in total? Explain with a series and with a shortcut.Show answerGames per round: 32, 16, 8, 4, 2, 1: geometric with a = 32, r = 1/2, n = 6: S₆ = 32(1 − (1/2)⁶) ÷ (1/2) = 64(63/64) = 63 games. Shortcut: every game eliminates exactly one team, and 63 of the 64 teams must be eliminated, so there are 63 games.Check your understandingQuiz1. Which question asks for a series?How many seats are in row 10?How high is the 3rd bounce?What is the 20th term?How many seats are in the first 10 rows altogether?2. A pile of logs has 12 on the bottom row, 11 on the next, and so on up to 1. How many logs?6678144723. A ball is dropped from 8 m and rebounds to 50% of its height each time. How far has it travelled when it hits the ground the 3rd time?14 m26 m20 m22 m4. Calls in a phone tree are 4, 16, 64, …. How many calls are made in the first 4 rounds?3402568410245. An object falls 4.9 m, 14.7 m, 24.5 m, … in successive seconds. How far does it fall in 5 seconds?44.1 m245 m24.5 m122.5 m6. Ella saves $10 the first week, then $3 more each week than the week before. Total after 10 weeks?$235$37$370$130Mark lesson completeFor parents and teachersBig idea: Expectation C2.4 asks students to solve problems involving arithmetic and geometric series, including real-world ones. Deciding between "one term" and "the total" is the main new skill, and C2.3's sum formulas are the tools.How to help: Ask your teen to underline the words that tell them it is a sum. For bouncing balls, insist on a sketch: the drop is counted once, and every rebound height twice.Common difficulties:Answering with a term when the question asks for a total, or the reverse.Bouncing-ball totals: forgetting to double the rebound heights, or counting the first drop twice.Counting the number of terms (rows, weeks, rounds) incorrectly. A short table of the first few values helps.Not checking whether the answer is reasonable (e.g., a total smaller than the largest term is impossible when all terms are positive).PreviousSolving Problems with Arithmetic and Geometric SequencesNextSequences and Series: Unit Review