DP Maths AI SL 1.2 Arithmetic Sequences and Series

DP Maths AI SL 1.2 Arithmetic Sequences and Series - Full notes

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DP Maths AI SL | 1.2 Arithmetic Sequences and Series

DP MATHEMATICS AI · TOPIC 1: NUMBER AND ALGEBRA

1.2

Arithmetic Sequences and Series

Full notes for Applications and Interpretation SL

Contents

1	Number patterns and sequences
2	Recursive and explicit definitions
3	Sigma notation
4	The nth term of an arithmetic sequence
5	Finding unknown terms, differences and positions
6	Arithmetic series
7	Sums with technology and inequalities in n
8	Arithmetic sequences as linear functions
9	Simple interest and depreciation by a fixed amount
10	Modelling real situations with arithmetic sequences
11	Deciding whether data are arithmetic
12	Problem solving with two sequences
13	Applications in science and everyday life
14	Spreadsheets and technology for sequences
15	Mixed review: worked problems
16	Investigation: staircases, stacks and triangular numbers
17	Exam-style questions with markscheme
18	Common misconceptions and links
19	Check your understanding
20	Summary and topic map
21	Answers to practice exercises

These are full teaching notes. Read each section in order, try the Think about it question before you read on, and work through every worked example before attempting the practice exercise.

Before you start

Rate yourself on each statement now, and again after you've worked through the notes. Shade R (not yet), A (getting there) or G (confident).

I can…	R	A	G
recognize patterns and describe them with recursive and explicit formulas
read, write and evaluate sigma notation
use un = u1 + (n − 1)d to find terms, the common difference and the position of a term
use Sn = n/2 (2u1 + (n − 1)d) and Sn = n/2 (u1 + un)
use technology (tables, solvers, sequence mode) to answer questions about n
interpret arithmetic sequences as linear models, including simple interest
decide whether real data are approximately arithmetic
solve problems involving two sequences

SL booklet: this booklet contains the content for Applications and Interpretation SL. Sections marked for HL in the HL edition are not included.

Guiding questions

How can patterns that grow by a constant amount be described and predicted? When is a linear (arithmetic) model appropriate for real data?

How this topic is examined

Arithmetic sequences appear on Paper 1 and Paper 2, very often in context: seats in a theatre, savings plans, salaries, training schedules and stacked objects. Questions ask for a term, a sum, the number of terms needed, or a comparison between two plans.

Write the formula you are using, show the values you substitute, and give integer answers where n must be a whole number. In context questions, finish with a sentence that answers the question asked.

Mathematics: Applications and Interpretation places strong emphasis on technology, modelling and interpretation. Most papers allow a GDC, so you are expected to use it efficiently, record the method you used, and interpret every answer in context.

Command term	What the examiner wants
Write down	an answer with little or no working; it should be obtainable by inspection
Find / Calculate / Determine	obtain the answer showing relevant working, including what you entered into a GDC
Show that	obtain the given result; every step must be shown and the result must not be assumed
Estimate	obtain an approximate value
Sketch	a diagram showing the general shape and key features, labelled
Interpret / Comment	explain the meaning of a result in context, or give a judgement
Justify / Explain	give valid reasons or evidence
Hence	use the previous result

How to use these notes

Box	What it's for
Think about it	a question to get you thinking before a section
Why it works	a derivation or explanation of a key result
Key point	a result or idea you must not forget
Common mistake	an error examiners see every year, and how to avoid it
GDC tip	how to use your graphic display calculator efficiently
In the real world	where the mathematics is used outside the classroom
Worked example	a full solution with IB-style marks (M1 method, A1 answer, R1 reasoning, AG answer given), graded basic, standard and exam level
Practice exercise	graded questions for each section; full answers are at the end of the booklet

1 Number patterns and sequences

Ordered lists that follow a rule.

Think about it

A gym membership costs $40 to join and $25 each month. Write down the total cost after 1, 2, 3 and 4 months. What do you notice about the differences between these totals?

A sequence is an ordered list of numbers. Each number is a term. We label the terms u1 (the first term), u2 (the second term), and in general un (the nth term). The subscript n is the position of the term, and it is always a positive integer.

Some sequences are finite (they stop), such as the number of seats in each of the 20 rows of a theatre. Others are infinite, such as the multiples of 7: 7, 14, 21, 28, ....

Sequences are everywhere in modelling. A sequence describes a quantity that is measured at regular steps: the balance of an account at the end of each year, the height of a plant at the end of each week, or the number of bricks in each layer of a wall. Studying how the terms change from step to step tells us which model to use.

In the gym example, the totals are 65, 90, 115, 140: each total is $25 more than the previous one. A sequence in which the difference between consecutive terms is constant is called arithmetic. This booklet is about such sequences and their sums.

Sequence	Rule in words	Arithmetic?
3, 7, 11, 15, ...	add 4 each time	yes, d = 4
50, 45, 40, 35, ...	subtract 5 each time	yes, d = −5
2, 6, 18, 54, ...	multiply by 3 each time	no (geometric)
1, 4, 9, 16, ...	square numbers	no
6.5, 6.5, 6.5, ...	constant	yes, d = 0

WORKED EXAMPLE 1 Continuing a pattern (basic)

Write down the next two terms and describe the rule: (a) 12, 19, 26, 33, ...; (b) 4.8, 4.2, 3.6, 3.0, ....

Solution

(a) Differences are 7: next terms 40, 47; add 7 each timeA1

(b) Differences are −0.6: next terms 2.4, 1.8; subtract 0.6 each timeA1

WORKED EXAMPLE 2 Testing for an arithmetic sequence (standard)

Determine whether 5, 8.5, 12, 15.5, 19 is arithmetic.

Solution

Differences: 3.5, 3.5, 3.5, 3.5M1

All equal, so arithmetic with d = 3.5A1

Practice exercise 1

1. Write down the next three terms of 101, 94, 87, ... .

2. Is 2, 5, 9, 14, ... arithmetic? Explain.

3. The first term of a sequence is 8 and each term is 3 less than the previous one. Write down the first five terms.

4. Which of these are arithmetic: (a) 1, 1.5, 2, 2.5; (b) 1, 2, 4, 8; (c) −4, −1, 2, 5?

2 Recursive and explicit definitions

Two ways to describe a sequence.

A recursive (or recurrence) definition gives the first term and a rule for getting each term from the previous one. For the gym example: u1 = 65, un+1 = un + 25. Recursive rules are natural for describing step-by-step processes, and they are exactly how spreadsheets and the GDC sequence mode generate terms.

An explicit definition gives the nth term directly as a function of n, for example un = 25n + 40. An explicit formula lets you jump straight to the 100th term without generating the 99 before it.

For an arithmetic sequence, the explicit formula is always linear in n: un = dn + c for constants d and c. The coefficient of n is the common difference.

WORKED EXAMPLE 3 From recursive to explicit (standard)

A sequence is defined by u1 = 7, un+1 = un + 4. Find an explicit formula and u30.

Solution

Terms 7, 11, 15, ...: arithmetic with d = 4M1

un = 7 + 4(n − 1) = 4n + 3A1

u30 = 123A1

WORKED EXAMPLE 4 From explicit to recursive (standard)

Given un = 18 − 2.5n, write a recursive definition.

Solution

u1 = 15.5A1

un+1 − un = −2.5, so un+1 = un − 2.5, u1 = 15.5A1

GDC tip

In sequence (or recursion) mode, enter u(n) = u(n − 1) + d with u(1) given, then view the table. Alternatively, enter the explicit formula y = dx + c as a function and use the table with x = 1, 2, 3, ....

Practice exercise 2

1. Write an explicit formula for u1 = −3, un+1 = un + 6.

2. Write a recursive definition for un = 9n − 4.

3. un = 120 − 7n. Find the first term that is negative.

3 Sigma notation

A compact way to write sums.

The capital Greek letter Σ (sigma) means 'sum'. The expression below means: substitute k = 1, then k = 2, and so on up to k = n, and add the results:

The letter k is a dummy variable: any letter may be used. The limits need not start at 1. Two useful rules are that constants can be taken outside the sum, and that the sum of a constant c over n terms is cn.

WORKED EXAMPLE 5 Evaluating sigma notation (basic)

Evaluate Σ from k = 1 to 5 of (3k + 2).

Solution

5 + 8 + 11 + 14 + 17M1

= 55A1

WORKED EXAMPLE 6 Writing sigma notation (standard)

Write 6 + 10 + 14 + ... + 86 in sigma notation.

Solution

uk = 4k + 2A1

4k + 2 = 86 gives k = 21M1

Σ from k = 1 to 21 of (4k + 2)A1

GDC tip

Most GDCs have a summation template (Σ) or a sum(seq(...)) command: sum(seq(3k + 2, k, 1, 5)) returns 55.

Practice exercise 3

1. Evaluate Σ from k = 3 to 8 of (10 − k).

2. How many terms are in Σ from r = 5 to 40?

3. Write 1.5 + 3 + 4.5 + ... + 30 in sigma notation.

4 The nth term of an arithmetic sequence

Start at u1, add d each step.

Starting from u1, the second term adds one d, the third adds two, and the nth adds (n − 1):

Here u1 is the first term, d the common difference (which may be negative or a decimal) and n the position. The formula is in the IB formula booklet, but you must be able to use it confidently in both directions.

Figure 1. un = 12 + 5(n − 1) plotted against n: the points lie on a straight line.

WORKED EXAMPLE 7 Finding a term (basic)

Find u40 for 12, 17, 22, ... .

Solution

u1 = 12, d = 5A1

u40 = 12 + 39(5) = 207A1

WORKED EXAMPLE 8 Decreasing sequence (standard)

The temperature in a cooling room falls by 1.8 °C each hour from 26 °C. Find the temperature after 9 hours.

Solution

Hour 0 is 26: u10 = 26 + 9(−1.8)M1

= 9.8 °CA1

Read the context carefully: the temperature 'after 9 hours' is 9 steps after the start, which is the 10th term if u1 is the starting value.

Common mistake

Using un = u1 + nd. Only (n − 1) differences are added to reach the nth term.

Practice exercise 4

1. Find the 25th term of 3, 9, 15, ... .

2. Find u60 for 100, 96.5, 93, ... .

3. A child saves $5 in week 1 and $2 more each week than the week before. How much is saved in week 12?

4. Find the 15th term of an arithmetic sequence with u1 = −4 and d = 0.75.

5 Finding unknown terms, differences and positions

Rearrange or solve simultaneously.

Questions often give two pieces of information and ask for the others. Typical situations: (1) two terms given: write each in the form u1 + (n − 1)d and solve simultaneously; (2) a term value given: solve u1 + (n − 1)d = value for n, which must be a positive integer; (3) three consecutive terms given in terms of x: use the fact that the middle term is the mean of its neighbours, 2b = a + c.

WORKED EXAMPLE 9 Two terms given (standard)

In an arithmetic sequence, u4 = 23 and u11 = 58. Find u1 and d.

Solution

u1 + 3d = 23; u1 + 10d = 58M1

7d = 35: d = 5A1

u1 = 8A1

WORKED EXAMPLE 10 Position of a term (standard)

Which term of 7, 13, 19, ... equals 307?

Solution

7 + 6(n − 1) = 307M1

n − 1 = 50A1

51st termA1

WORKED EXAMPLE 11 Is it a term? (exam level)

Determine whether 500 is a term of 4, 15, 26, ... .

Solution

4 + 11(n − 1) = 500 gives n − 1 = 45.09...M1 A1

n is not an integer, so 500 is not a termR1

WORKED EXAMPLE 12 Three consecutive terms (exam level)

x + 3, 3x − 1 and 2x + 8 are consecutive terms of an arithmetic sequence. Find x.

Solution

2(3x − 1) = (x + 3) + (2x + 8)M1

6x − 2 = 3x + 11A1

x = 13/3A1

Practice exercise 5

1. u3 = 14 and u9 = 50. Find u1, d and u20.

2. Which term of 200, 192, 184, ... is 0?

3. Is 1000 a term of 3, 10, 17, ...?

4. Find k if k, 2k + 1 and 5k − 3 are consecutive terms of an arithmetic sequence.

5. The 6th term is 3 times the 2nd term, and the 10th term is 52. Find u1 and d.

6 Arithmetic series

Adding up the terms.

Think about it

Without a calculator, find 1 + 2 + 3 + ... + 100. Hint: pair the first and last terms.

An arithmetic series is the sum of the terms of an arithmetic sequence. Sn denotes the sum of the first n terms. Writing the sum forwards and backwards and adding pairs gives:

Why it works

Sn = u1 + (u1 + d) + ... + un. Written backwards, Sn = un + (un − d) + ... + u1. Adding term by term, each of the n pairs sums to u1 + un, so 2Sn = n(u1 + un). Substituting un = u1 + (n − 1)d gives the second form. For 1 + 2 + ... + 100: 50 pairs of 101 give 5050.

Use the first form when you know the last term, and the second when you know d. Note that un = Sn − Sn−1: the nth term is the difference between two consecutive sums.

WORKED EXAMPLE 13 Sum of terms (basic)

Find the sum of the first 30 terms of 5, 9, 13, ... .

Solution

S30 = 15(10 + 29 × 4)M1

= 1890A1

WORKED EXAMPLE 14 Sum with a known last term (standard)

Find 20 + 23 + 26 + ... + 140.

Solution

140 = 20 + 3(n − 1): n = 41M1 A1

S41 = 41/2 × (20 + 140) = 3280A1

WORKED EXAMPLE 15 From a sum to the terms (exam level)

The sum of the first n terms of a sequence is Sn = 3n2 + 2n. Find u1, u2 and show the sequence is arithmetic.

Solution

u1 = S1 = 5; S2 = 16 so u2 = 11A1 A1

un = Sn − Sn−1 = 6n − 1, linear in n, so arithmetic with d = 6M1 R1

Practice exercise 6

1. Find the sum of the first 50 even numbers.

2. Find 100 + 97 + 94 + ... + 1.

3. An arithmetic series has u1 = 8 and S12 = 294. Find d.

4. Sn = 2n2 − n. Find u10.

5. Find the sum of all multiples of 9 between 100 and 500.

7 Sums with technology and inequalities in n

When n is the unknown.

A common question asks for the smallest n for which a sum exceeds a target. Set up the inequality Sn > target. This is a quadratic inequality in n. On Paper 2, solve it with a GDC table, a graph, or the polynomial solver, and then round up to the next integer, checking the neighbouring values.

WORKED EXAMPLE 16 Smallest number of terms (standard)

How many terms of 4 + 7 + 10 + ... are needed for the sum to exceed 2000?

Solution

n/2 (8 + 3(n − 1)) > 2000, so 3n2 + 5n − 4000 > 0M1

Positive root n = 35.68A1

36 terms (S35 = 1960, S36 = 2088)A1

WORKED EXAMPLE 17 Using a table (exam level)

A runner runs 3 km on day 1 and increases by 0.4 km each day. On which day does her total distance first exceed 200 km?

Solution

Sn = n/2 (6 + 0.4(n − 1))M1

Table: S25 = 195, S26 = 208A1

Day 26A1

Checking the table values on either side of the answer confirms it.

GDC tip

Enter y = x/2 (6 + 0.4(x − 1)) and scroll the table from x = 20. Record the two values either side of the target in your answer.

Practice exercise 7

1. Find the smallest n for which 2 + 5 + 8 + ... exceeds 1000.

2. How many terms of 50 + 48 + 46 + ... give a total of 650?

3. Find the greatest n for which Sn of 100, 95, 90, ... is positive.

8 Arithmetic sequences as linear functions

Discrete points on a straight line.

Since un = d n + (u1 − d), the terms of an arithmetic sequence lie on a straight line with gradient d and y-intercept u1 − d (which would be the 'zeroth term'). The difference from a linear function is that a sequence is discrete: it is only defined for n = 1, 2, 3, ..., so its graph is a set of separate points.

This connection explains why arithmetic sequences model linear growth: equal increases in equal time steps.

Figure 2. Arithmetic sequences with positive and negative common differences.

WORKED EXAMPLE 18 From a graph (standard)

The points (1, 7), (2, 10.5), (3, 14), ... lie on a line. Write down d and un.

Solution

d = gradient = 3.5A1

un = 3.5n + 3.5A1

Practice exercise 8

1. A sequence has un = 4n − 9. State the gradient of the line through its points and the 'zeroth term'.

2. Explain why the graph of an arithmetic sequence should not be drawn as a continuous line.

9 Simple interest and depreciation by a fixed amount

Linear financial models.

With simple interest, interest is paid only on the original amount (the principal P). At r% per year, the interest each year is Pr/100, so the balance increases by the same amount each year: an arithmetic sequence. The total interest after n years is I = Prn/100.

Similarly, straight-line depreciation reduces the value of an asset by the same amount each year. (Depreciation by a fixed percentage is geometric; see the geometric sequences booklet.)

WORKED EXAMPLE 19 Simple interest (basic)

$6000 is invested at 3.5% simple interest per year. Find the balance after 8 years.

Solution

I = 6000 × 3.5 × 8/100 = 1680M1

Balance $7680A1

WORKED EXAMPLE 20 Straight-line depreciation (standard)

A machine costs $48 000 and loses $5500 in value each year. After how many full years is its value below $10 000?

Solution

48 000 − 5500n < 10 000M1

n > 6.9A1

7 yearsA1

Practice exercise 9

1. Find the simple interest on $2500 at 4% for 6 years.

2. How many years does it take $1200 to grow to $1500 at 5% simple interest?

3. A laptop worth $1800 loses $250 each year. Find its value after 5 years and when it is first worth less than $300.

10 Modelling real situations with arithmetic sequences

Rows, layers, schedules.

Ask: does the quantity change by the same amount in each step? If yes, an arithmetic model fits. Identify u1, d and what n represents, and decide whether the question asks for a single term (un) or a total (Sn).

WORKED EXAMPLE 21 Theatre seating (standard)

A theatre has 24 seats in the front row and each row has 2 more seats than the row in front. There are 30 rows. Find the seats in the last row and the total number of seats.

Solution

u30 = 24 + 29(2) = 82A1

S30 = 15(24 + 82) = 1590M1 A1

WORKED EXAMPLE 22 Stack of logs (standard)

Logs are stacked with 20 in the bottom layer, 19 in the next, and so on, with 1 at the top. How many logs are there?

Solution

S20 = 20/2 (20 + 1)M1

= 210A1

WORKED EXAMPLE 23 Salary comparison (exam level)

Job A starts at $42 000 and rises by $1800 per year. Job B starts at $38 000 and rises by $2600 per year. In which year does B's salary first exceed A's, and which job pays more in total over 10 years?

Solution

38 000 + 2600(n − 1) > 42 000 + 1800(n − 1): 800(n − 1) > 4000, n > 6M1 A1

Year 7A1

Totals over 10 years: A = 5(84 000 + 16 200) = $501 000; B = 5(76 000 + 23 400) = $497 000: A pays moreM1 A1

Practice exercise 10

1. A clock strikes once at 1 o'clock, twice at 2 o'clock, and so on. How many strikes are there in 12 hours?

2. A worker earns $15 per hour in year 1, increasing by $0.60 per hour each year. In which year does the rate first exceed $20?

3. A training plan starts with 20 push-ups and adds 3 each day. Find the total in 4 weeks.

11 Deciding whether data are arithmetic

Real data are rarely perfect.

In Applications and Interpretation, models are used for real data that only approximately follow a pattern. To decide whether an arithmetic model is suitable, look at the differences between consecutive values: if they are roughly constant, a linear model is appropriate. The average difference gives an estimate of d.

When data are noisy, linear regression (see the statistics booklets) gives a better estimate. Always comment on how well the model fits and on the danger of extrapolating far beyond the data.

Year	2019	2020	2021	2022	2023
Members	412	455	496	541	583
Difference		43	41	45	42

WORKED EXAMPLE 24 Fitting an arithmetic model (exam level)

Using the table, estimate d, write a model for the number of members in year n (n = 1 for 2019), and predict 2030.

Solution

Average difference = (583 − 412)/4 = 42.75M1 A1

un = 412 + 42.75(n − 1)A1

2030 is n = 12: 882; extrapolation beyond the data, so treat with cautionA1 R1

Practice exercise 11

1. The heights of a plant (cm) at the end of weeks 1 to 5 are 4.1, 6.0, 8.2, 9.9, 12.1. Is an arithmetic model reasonable? Estimate d.

2. Explain why an arithmetic model for a plant's height cannot be valid indefinitely.

12 Problem solving with two sequences

Compare, equate and combine.

Problems with two sequences ask when one overtakes the other (set un for one greater than un for the other) or when their sums are equal. Because both are linear in n, comparisons of terms lead to linear inequalities, while comparisons of sums lead to quadratic inequalities.

WORKED EXAMPLE 25 When are the sums equal? (exam level)

Sequence A: 30, 32, 34, ...; sequence B: 10, 15, 20, .... Find n for which the sums of the first n terms are equal.

Solution

n/2 (60 + 2(n − 1)) = n/2 (20 + 5(n − 1))M1

58 + 2n = 15 + 5nA1

n = 43/3: not an integer, so the sums are never exactly equal; B's sum first exceeds A's at n = 15R1 A1

Practice exercise 12

1. Two arithmetic sequences have un = 5n + 12 and vn = 8n − 3. Find the first n with vn > un.

2. Two savings plans: Plan P saves $100 in month 1 rising by $10 each month; plan Q saves $150 each month. After how many months does P's total first exceed Q's?

13 Applications in science and everyday life

Linear patterns in the world around us.

Arithmetic sequences appear wherever a quantity changes by a fixed amount at regular intervals: a car's odometer reading on a daily commute of fixed length; the temperature of rock increasing by roughly 25 °C for each kilometre of depth; taxi fares with a fixed charge per kilometre; the dates of events that repeat every week; and the radius of tree rings in years of steady growth.

In science, constant-difference patterns often indicate constant rates: a car moving at constant velocity covers equal distances in equal times; a tank filled at a constant rate gains equal volumes each minute. Recognizing the constant difference is the first step to choosing a linear model.

WORKED EXAMPLE 26 Geothermal gradient (standard)

Rock temperature is 15 °C at the surface and increases by 25 °C per km of depth. At what depth is the temperature 140 °C?

Solution

15 + 25k = 140M1

k = 5 kmA1

WORKED EXAMPLE 27 Calendar patterns (standard)

A club meets every 9 days, first on 3 January (day 3 of the year). On which day of the year is the 30th meeting, and how many meetings are there in a 365-day year?

Solution

Day of meeting n: 3 + 9(n − 1); n = 30 gives day 264M1 A1

3 + 9(n − 1) ≤ 365 gives n ≤ 41.2: 41 meetingsM1 A1

WORKED EXAMPLE 28 Filling a tank (standard)

A 900-litre tank already contains 120 litres and is filled at 7.5 litres per minute. Write the volume after n minutes and find when the tank is 80% full.

Solution

Vn = 120 + 7.5nA1

120 + 7.5n = 720 gives n = 80 minutesM1 A1

Always identify the starting value and the step size from the context before writing a model.

Practice exercise 13

1. A taxi charges $3.20 plus $1.85 per km. Find the fare for 12 km and the distance for a fare of $40.20.

2. A runner's training distance increases by 0.5 km each week from 5 km. In which week does she first run a half marathon (21.1 km)?

3. Seats in a stadium section increase by 4 per row from 22. How many rows are needed for at least 1000 seats?

14 Spreadsheets and technology for sequences

Generate, tabulate and graph.

Spreadsheets are the natural tool for sequences in Applications and Interpretation. Put n in column A (1, 2, 3, ...), the first term in B1, and the recursive rule =B1+d in B2, filled down. A running total in column C (=C1+B2) gives Sn. Charts of columns B and C show the linear growth of un and the quadratic growth of Sn.

On a GDC, the same three tools exist: sequence mode (recursive), function tables (explicit formula), and the list/statistics editor (seq and cumSum commands). Choose whichever lets you show your method most clearly.

n	un (=B1+3)	Sn (running total)
1	5	5
2	8	13
3	11	24
4	14	38
5	17	55
6	20	75

Figure 3. Terms grow linearly while sums grow quadratically.

WORKED EXAMPLE 29 Running totals in a list (standard)

Use cumSum(seq(5 + 3(k − 1), k, 1, 20)) to find S20 and the first n with Sn > 300.

Solution

The list of running totals ends with S20 = 670A1

Scanning the list: S13 = 299, S14 = 343, so n = 14A1

Practice exercise 14

1. Set up a spreadsheet for u1 = 12, d = −1.5 and find the first negative term.

2. Use a running total to find the first n for which 2 + 9 + 16 + ... exceeds 500.

15 Mixed review: worked problems

Bringing the ideas together.

WORKED EXAMPLE 30 Mixed problem 1 (standard)

The 3rd term of an arithmetic sequence is 11 and the sum of the first 10 terms is 185. Find u1 and d.

Solution

u1 + 2d = 11; 5(2u1 + 9d) = 185M1

2u1 + 9d = 37; subtract 2(u1 + 2d) = 22: 5d = 15A1

d = 3, u1 = 5A1

WORKED EXAMPLE 31 Mixed problem 2 (standard)

A ladder has 15 rungs whose lengths decrease uniformly from 45 cm at the bottom to 31 cm at the top. Find the total length of wood used for the rungs.

Solution

S15 = 15/2 (45 + 31)M1

= 570 cmA1

WORKED EXAMPLE 32 Mixed problem 3 (exam level)

The sum of the first n terms of an arithmetic sequence is 4n2 + n. Find the first term, the common difference and the 25th term.

Solution

u1 = S1 = 5A1

u2 = S2 − S1 = 18 − 5 = 13, d = 8A1

u25 = 5 + 24 × 8 = 197A1

WORKED EXAMPLE 33 Mixed problem 4 (exam level)

A loan of $9000 is repaid by monthly payments that increase by $20 each month, starting with $150. How many months are needed to repay the loan (ignoring interest)?

Solution

n/2 (300 + 20(n − 1)) ≥ 9000M1

10n2 + 140n − 9000 ≥ 0: n ≥ 23.8A1

24 monthsA1

Practice exercise 15

1. Find the number of terms in 7, 11, 15, ..., 403, and their sum.

2. u5 + u9 = 72 and u7 × u2 = 90. Find the possible values of d.

3. A stadium has 40 seats in the first row and 3 more in each subsequent row, with a total of 5304 seats. How many rows are there?

16 Investigation: staircases, stacks and triangular numbers

A Paper 3 style exploration.

Triangular numbers 1, 3, 6, 10, 15, ... count the dots in triangular arrangements. They are not arithmetic, but they are the sums of an arithmetic sequence: Tn = 1 + 2 + ... + n = n(n + 1)/2.

Work through the following investigation, which mirrors the structure of an extended Paper 3 question.

1. Show that Tn = n(n + 1)/2 using the arithmetic series formula. 2. Show that Tn + Tn−1 = n2, and illustrate this with dot diagrams. 3. A staircase is built from cubes: the first step uses 1 cube, the second 2, and so on. How many cubes are needed for a 20-step staircase? 4. A pyramid of cans has 1 can on top, then 3, then 5, ... (odd numbers). How many layers can be built with 400 cans? 5. Investigate which triangular numbers are also square numbers.

Answers. 1. Sn = n/2 (1 + n). 2. n(n + 1)/2 + (n − 1)n/2 = n2. 3. T20 = 210. 4. 1 + 3 + 5 + ... = n2 ≤ 400, so 20 layers. 5. 1, 36, 1225, ... (T1, T8, T49).

17 Exam-style questions with markscheme

Attempt each question under timed conditions before reading the markscheme.

Question 1 (Paper 1 style) (about 10 minutes)

The first three terms of an arithmetic sequence are 2k + 1, 5k − 2 and 6k + 3.

(a) Find k. [2]

(b) Find the common difference. [1]

Markscheme

Part	Answer	Marks
(a)	2(5k − 2) = 2k + 1 + 6k + 3	M1
	k = 4	A1
(b)	terms 9, 18, 27: d = 9	A1
(c)	S20 = 10(18 + 19 × 9)	M1 A1
	1890	A1

Examiner's comments. Using 2b = a + c gives k directly.

Question 2 (Paper 2 style) (about 15 minutes)

A company sells 800 units in its first month. Sales increase by 35 units each month.

(a) Find the number of units sold in month 24. [2]

(b) Find the total units sold in the first two years. [2]

(d) Find the first month by the end of which total sales exceed 60 000 units. [3]

Markscheme

Part	Answer	Marks
(a)	800 + 23 × 35	M1
	1605	A1
(b)	12(1600 + 23 × 35)	M1
	28 860	A1
(c)	800 + 35(n − 1) > 2000	M1
	n > 35.3	A1
	month 36	A1
(d)	n/2 (1600 + 35(n − 1)) > 60 000	M1
	n > 40.3 (S40 = 59 300, S41 = 61 500)	A1
	month 41	A1

Examiner's comments. Integer answers must be justified from the inequality or a table.

Question 3 (Paper 2 style) (about 12 minutes)

Theatre A has 18 seats in row 1, increasing by 2 seats per row. Theatre B has 30 seats in row 1, increasing by 1 seat per row. Both have n rows.

(a) Write expressions for the total seats in each theatre. [2]

(b) Find the least n for which theatre A has more seats than theatre B. [3]

Markscheme

Part	Answer	Marks
(a)	A: n(18 + n − 1) = n2 + 17n	A1
	B: n/2 (60 + n − 1) = (n2 + 59n)/2	A1
(b)	n2 + 17n > (n2 + 59n)/2	M1
	n2 > 25n, n > 25	A1
	n = 26	A1
(c)	26 rows is plausible for a large theatre; very wide rows may be impractical	R1

Examiner's comments. Reasonable comments in context are credited.

18 Common misconceptions and links

Students often think…	Actually…
un = u1 + nd	un = u1 + (n − 1)d
Sn is the nth term	Sn is the sum of n terms
n can be 35.7	n must be a positive integer
straight-line depreciation is geometric	it is arithmetic

How 1.2 connects to the rest of the course

Topic	Connection
1.3 Geometric sequences	constant ratio
2.2 Linear functions	gradient and intercept
4.4 Regression	fitting linear models
1.4 Finance	simple and compound interest

19 Check your understanding

Section A: multiple choice

1. The common difference of 15, 11, 7, ... is

A. −4 B. 4 C. −3 D. 11

2. u10 of 2, 5, 8, ... is

A. 30 B. 32 C. 29 D. 27

3. S4 of 3, 7, 11, 15 is

A. 15 B. 36 C. 30 D. 48

4. An arithmetic sequence plotted against n lies on

A. a parabola B. an exponential curve C. a circle D. a straight line

5. Simple interest balances form

A. a geometric sequence B. an arithmetic sequence C. a constant sequence D. a quadratic sequence

Section B: structured

6. Find the sum of the integers from 1 to 200 that are not multiples of 4. [4]

7. Explain why Sn for an arithmetic series is a quadratic in n. [2]

Answers

1. A 2. C 3. B 4. D 5. B

6. 20 100 − 4(1 + ... + 50) = 20 100 − 5100 = 15 000.

7. Sn = (d/2)n2 + (u1 − d/2)n.

20 Summary and topic map

Idea	Remember
Arithmetic sequence	constant difference d
nth term	un = u1 + (n − 1)d
Sum	Sn = n/2 (2u1 + (n − 1)d) = n/2 (u1 + un)
Sigma	Σ; number of terms = upper − lower + 1
Linear model	gradient d; discrete points
Simple interest	I = Prn/100
Technology	tables and solvers for n

Before the exam, make sure you…

•identify u1, d and n in context
•round n up when a target must be exceeded
•check data differences before using a model
•answer in context

Arithmetic Sequences and Series on one page

Cover this page with a sheet of paper and try to rebuild it from memory. Whatever you can't recall is what to revise next.

Figure 4. Topic map for Arithmetic Sequences and Series.

21 Answers to practice exercises

Check your working against these answers. Full methods are expected in the exam.

Practice exercise 1

1. 80, 73, 66

2. no: differences 3, 4, 5 are not constant

3. 8, 5, 2, −1, −4

4. (a) and (c)

Practice exercise 2

1. un = 6n − 9

2. u1 = 5, un+1 = un + 9

3. u18 = −6

Practice exercise 3

1. 27

2. 36

3. Σ from k = 1 to 20 of 1.5k

Practice exercise 4

1. 147

2. −96.5

3. $27

4. 6.5

Practice exercise 5

1. u1 = 2, d = 6, u20 = 116

2. 26th

3. no: n − 1 = 142.4 is not an integer

4. 2(2k + 1) = 6k − 3: k = 2.5

5. u1 + 5d = 3(u1 + d) gives u1 = d; 10d = 52, d = 5.2 = u1

Practice exercise 6

1. 2550

2. 34 terms: 1717

3. d = 3

4. 37

5. 108 to 495: 44 terms, S = 13 266

Practice exercise 7

1. n = 26

2. n2 − 51n + 650 = 0: n = 25 or 26

3. Sn = n/2 (205 − 5n) > 0: n = 40

Practice exercise 8

1. 4; −9

2. n takes only integer values

Practice exercise 9

1. $600

2. 5 years

3. $550; after 7 years ($50)

Practice exercise 10

1. 78

2. 15 + 0.6(n − 1) > 20: n = 10

3. 28 days: 14(40 + 81) = 1694

Practice exercise 11

1. differences 1.9, 2.2, 1.7, 2.2: roughly constant; d ≈ 2.0

2. growth slows and stops; heights cannot increase without limit

Practice exercise 12

1. n = 6

2. n/2 (200 + 10(n − 1)) > 150n gives n > 11: month 12

Practice exercise 13

1. $25.40; 20 km

2. 5 + 0.5(n − 1) ≥ 21.1: week 34

3. n/2 (44 + 4(n − 1)) ≥ 1000: 2n2 + 20n − 1000 ≥ 0, n = 18

Practice exercise 14

1. u10 = −1.5

2. S12 = 486, S13 = 572: n = 13

Practice exercise 15

1. 100 terms; 20 500

2. u7 = 36 so u2 = 2.5: d = 6.7

3. n/2 (80 + 3(n − 1)) = 5304: 3n2 + 77n − 10 608 = 0, n = 48

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