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Arithmetic sequences in real life

Patterns and sequences · level 5, lesson 6 of 6

A question to start

A shop wants to stack 100 cans into a triangle for a display: one can on top, two under it, three under those, and so on, each row one longer than the row above. How many rows can they make? Will any cans be left over?

You could stack real cans and count. Or you could notice that the rows form an arithmetic sequence and that the total is one of Gauss's sums, and answer it with a pencil in a minute. This lesson is about spotting arithmetic sequences hiding in everyday situations, and about the small traps that catch people when they do.

The idea

When is a situation arithmetic?

A real situation can be modelled by an arithmetic sequence when something changes by the same amount at each step. Clues in the wording: "each row has 2 more seats", "she saves 15 dollars every week", "the fee goes up by 3 dollars per kilometre", "each step is 18 cm higher". The step can be a row, a week, a kilometre or a step of a staircase; what matters is that the steps are equal and the change at every step is the same.

If the change is a percentage ("grows by 5% a year", "loses half its value"), the situation is not arithmetic. Those are multiplying patterns, the geometric sequences of level 6.

Four questions to ask before calculating

  1. What are \(a\) and \(d\)? What is the first value, and how much does it change each step? Is \(d\) positive or negative?
  2. What does \(n\) count, and where does it start? Is the first row \(n = 1\)? Or does the story start at "week 0", before anything has happened? Starting at 0 changes the formula: with a start value \(s\) at step 0, the value after \(n\) steps is \(s + nd\), not \(s + (n - 1)d\).
  3. Am I asked for one term or a total? "How many seats in the last row?" is a term, \(t_n\). "How many seats in the theatre?" is a sum, \(S_n\). Reading this wrong is the most common modelling mistake of all.
  4. Does the answer make sense? Rows cannot hold 12.4 people, and a savings total cannot be smaller than the last deposit. A rough estimate first ("about 25 rows of about 40 seats, so about 1000") catches big slips.

Stacking things: the rows and the total

Stacks are a natural home for arithmetic sequences, because each layer is one shorter (or one longer) than the layer below:

row 1: 5 cans row 2: 4 cans row 3: 3 cans row 4: 2 cans row 5: 1 can 5 + 4 + 3 + 2 + 1 = 15

The rows from the bottom are \(5, 4, 3, 2, 1\), an arithmetic sequence with \(d = -1\), and the whole stack is their sum. Reading from the top they are \(1, 2, 3, 4, 5\), and the total of \(n\) rows is the triangular number \(\dfrac{n(n + 1)}{2}\).

The fence-post trap, once more

A fence 30 metres long has a post every 2.5 metres, including both ends. How many posts? \(30 \div 2.5 = 12\) is the number of gaps; the number of posts is 13. The positions of the posts, 0, 2.5, 5, ..., 30, form an arithmetic sequence, and the general term \(0 + (n - 1) \times 2.5 = 30\) gives \(n = 13\) directly. Many real problems (posts, floors of a building, birthdays between two dates, pages from page 12 to page 40) hide this trap. The formula with \(n - 1\) in it never falls into it.

Worked examples

1. Theatre seats. The first row of a theatre has 18 seats and each row has 2 more than the row in front. There are 25 rows. How many seats in the last row, and in the whole theatre?

  • \(a = 18\), \(d = 2\), \(n = 25\).
  • Last row (a term): \(t_{25} = 18 + 24 \times 2 = 66\) seats.
  • Whole theatre (a sum): \(S_{25} = \dfrac{25 \times (18 + 66)}{2} = \dfrac{25 \times 84}{2} = 1050\) seats.
  • Estimate: 25 rows averaging about 42 seats is about 1000. ✓

2. A savings plan. You save 5 dollars in week 1, 7 dollars in week 2, 9 dollars in week 3, always 2 dollars more than the week before. How much do you put in during week 52, and how much have you saved in a year?

  • Week 52: \(t_{52} = 5 + 51 \times 2 = 107\) dollars.
  • Total: \(S_{52} = \dfrac{52 \times (5 + 107)}{2} = 26 \times 112 = 2912\) dollars.

Compare with a flat plan of 56 dollars every week: \(52 \times 56 = 2912\) dollars, exactly the same. That is no accident: 56 is the average of 5 and 107, and Gauss's formula says the total is "number of weeks times the average deposit".

3. Two salary offers. Job A pays 40 000 dollars in year 1 with a raise of 2000 dollars every year. Job B pays 45 000 dollars in year 1 with a raise of 1000 dollars every year.

  • Yearly salaries: A is \(40\,000 + 2000(n - 1)\) and B is \(45\,000 + 1000(n - 1)\). They are equal when \(1000(n - 1) = 5000\), so \(n - 1 = 5\): in year 6 both pay 50 000 dollars. After that A pays more.
  • Total over 10 years: A earns \(\dfrac{10}{2}\big[\,80\,000 + 9 \times 2000\,\big] = 490\,000\) dollars; B earns \(\dfrac{10}{2}\big[\,90\,000 + 9 \times 1000\,\big] = 495\,000\) dollars. B is still ahead in total.
  • Over 20 years: A earns \(10 \times (80\,000 + 38\,000) = 1\,180\,000\) dollars; B earns \(10 \times (90\,000 + 19\,000) = 1\,090\,000\) dollars. A is well ahead.

The salaries cross in year 6, but the totals cross much later, because B built up a lead in the early years. The question "which is better?" depends on how long you stay.

4. The can display. With \(n\) rows the stack holds \(\dfrac{n(n + 1)}{2}\) cans. For 13 rows that is \(\dfrac{13 \times 14}{2} = 91\); for 14 rows it is \(\dfrac{14 \times 15}{2} = 105\), too many. So 100 cans make 13 full rows with 9 cans left over.

5. Trips to Mars. Spacecraft are usually sent from Earth to Mars when the two planets line up well, which happens about once every 26 months. If a launch window opens in some month (call it month 0), the windows open in months 0, 26, 52, 78, ... The 10th window after this one is \(10 \times 26 = 260\) months later, about 21 years and 8 months. (The planets' orbits are ovals, not circles, so real windows drift by a few weeks; 26 months is the average spacing.) Here the story starts at "window 0", so the formula is \(26n\) rather than \(26(n - 1)\): question 2 of the four.

Try it

  1. A hall has 20 rows. The front row has 12 chairs and each row has 3 more than the one in front. How many chairs in the back row? In the hall?
  2. Logs are piled with 15 on the bottom layer, 14 on the next, and so on up to 1 on top. How many logs?
  3. A taxi charges 4 dollars to start plus 1.75 dollars for each kilometre. Write the cost of a trip of \(k\) kilometres, and find the cost of a 12 km trip.
  4. A runner trains by running 2 km on day 1 and 0.5 km more each day. On which day does she first run 10 km? How far has she run in total by the end of that day?
  5. A triangular can display uses exactly 120 cans. How many rows does it have?
  6. A fence 30 m long has a post every 2.5 m, including both ends. How many posts? What if the fence goes all the way around a field with a perimeter of 30 m?
  7. Plan A: you start with 100 dollars and add 10 dollars each week. Plan B: you start with nothing and add 25 dollars each week. After how many weeks does B first have more?
  8. A theatre has 30 rows and 1530 seats. Each row has 2 more seats than the row in front. How many seats are in the front row?
Answers
  1. Back row: \(12 + 19 \times 3 = 69\) chairs. Hall: \(\dfrac{20 \times (12 + 69)}{2} = 10 \times 81 = 810\) chairs.
  2. \(15 + 14 + \cdots + 1 = \dfrac{15 \times 16}{2} = 120\) logs.
  3. Cost \(= 4 + 1.75k\) dollars (here the start, \(k = 0\), already costs 4 dollars). For 12 km: \(4 + 21 = 25\) dollars.
  4. \(2 + (n - 1) \times 0.5 = 10\) gives \((n - 1) \times 0.5 = 8\), so \(n - 1 = 16\) and \(n = 17\): day 17. Total: \(\dfrac{17 \times (2 + 10)}{2} = 102\) km.
  5. \(\dfrac{n(n + 1)}{2} = 120\), so \(n(n + 1) = 240 = 15 \times 16\). 15 rows.
  6. 12 gaps of 2.5 m, so 13 posts in a straight line. Around a closed loop the last gap ends at the first post, so there are as many posts as gaps: 12 posts.
  7. After \(n\) weeks A has \(100 + 10n\) dollars and B has \(25n\) dollars. B is ahead when \(25n > 100 + 10n\), that is \(15n > 100\), \(n > 6.67\). So after 7 weeks (B has 175 dollars, A has 170 dollars). After 6 weeks A still leads, 160 dollars to 150 dollars.
  8. \(S_{30} = \dfrac{30}{2}\big[\,2a + 29 \times 2\,\big] = 15(2a + 58) = 1530\). So \(2a + 58 = 102\), \(2a = 44\), \(a = 22\) seats in the front row. (Check: back row \(22 + 58 = 80\), and \(\dfrac{30 \times (22 + 80)}{2} = 1530\). ✓)

Common mistake

Answering a term question with a sum, or the other way round. "How much do you save in week 52?" is 107 dollars; "how much have you saved after 52 weeks?" is 2912 dollars. Underline the question before you start.

Forcing an arithmetic model. A plant that grew 2 cm last week will not grow 2 cm every week forever, and a savings account that earns a percentage is not arithmetic. Use an arithmetic sequence only when the story really says "the same amount each step".

Where it leads