Module 7 of 46Lesson 2 of 10 · AA 1.2 — Arithmetic sequences and series

Course map · Topic 1: Number and Algebra

1.2 Arithmetic sequences and series

SL & HL12 worked questions

An arithmetic sequence changes by the same amount every step. In AA you must find terms and sums by hand on Paper 1 (often with algebra and a parameter $k$ ), and solve harder problems with the GDC on Paper 2.

✏️ Start with the questions. Try each one on paper, then open the solution and compare.

Stuck? Open the notes ↓

✏️ Practice questions

Try each question on paper first. Then open the solution and compare every step.

Question 1●○○ Warm-up✍️ By hand · Paper 1 · no calculator[5 marks]

Consider the arithmetic sequence

5,\ 9,\ 13,\ 17,\ \ldots

(a) Write down the common difference.[1]
(b) Find $u_{25}$ .[2]
(c) Find $S_{25}$ .[2]

Show the full step-by-step solution

Step 1🧠 ThinkWhat is being asked?

u_{25}

is the 25th number in the list.

S_{25}

is the total of the first 25 numbers.

Part (a)

Step 1✍️ By hand

d = 9 - 5 = 4

Part (b)

Step 1✍️ By hand24 jumps of 4 after the first term

u_{25} = 5 + 24\times4

= 101

Part (c)

Step 1✍️ By handWe know the first and last terms

S_{25} = \dfrac{25}{2}\left(5 + 101\right)

= \dfrac{25}{2}\times106 = 25\times53

= 1325

✅ Check it At home you can check this on the GDC: type sum(seq(5+4·(n−1),n,1,25)) — it gives 1325.

Final answer (a)

4

(b)

101

(c)

1325

Question 2●○○ Warm-up✍️ By hand · Paper 1 · no calculator[4 marks]

An arithmetic sequence has

u_1 = 40

and

d = -3

. Find the first negative term of the sequence, and state its position.

Show the full step-by-step solution

Step 1🧠 ThinkTurn “negative” into an inequality

The terms go down by 3 each time. We want the first

n

where

u_n \lt 0

Step 2✍️ By handSolve the inequality

40 - 3(n-1) \lt 0

40 \lt 3(n-1)

n - 1 \gt 13.\overline{3}

n \gt 14.\overline{3}\ \Rightarrow\ n = 15

u_{15} = 40 - 3\times14 = -2

A1A1

✅ Check it

u_{14} = 40 - 39 = 1

(still positive) and

u_{15} = -2

(negative). ✔

Final answer

u_{15} = -2

⚠️ Common mistake Dividing by a negative and forgetting to flip the inequality. Keeping the 3 positive, as above, avoids this.

Question 3●○○ Warm-up✍️ By hand · Paper 1 · no calculator[4 marks]

Find the value of

\displaystyle\sum_{k=1}^{20}(3k+2)

Show the full step-by-step solution

Step 1🧠 ThinkRead the sigma

Put

k = 1, 2, 3, \ldots, 20

into

3k+2

: we get

5, 8, 11, \ldots

. These go up by 3, so it is an arithmetic series with 20 terms.

Step 2✍️ By handFirst term, last term, sum

u_1 = 3(1)+2 = 5,\quad u_{20} = 3(20)+2 = 62

S_{20} = \dfrac{20}{2}\left(5 + 62\right)

= 10\times67

= 670

💡 Remember The number in front of

k

is the common difference:

3k+2

has

d = 3

Final answer

670

Question 4●●○ Standard✍️ By hand · Paper 1 · no calculator[5 marks]

In an arithmetic sequence,

u_4 = 17

and

u_{11} = 45

(a) Find the common difference and the first term.[3]
(b) Find $S_{20}$ .[2]

Show the full step-by-step solution

Step 1🧠 ThinkTwo facts → two equations

u_4

is 3 jumps after

u_1

;

u_{11}

is 10 jumps after

u_1

. From

u_4

u_{11}

there are 7 jumps.

Part (a)

Step 1✍️ By hand

u_1 + 3d = 17,\quad u_1 + 10d = 45

Subtract:

7d = 28

d = 4

u_1 = 17 - 12 = 5

Part (b)

Step 1✍️ By hand

S_{20} = \dfrac{20}{2}\left(2\times5 + 19\times4\right)

= 10\times86

= 860

Final answer (a)

d = 4,\ u_1 = 5

(b)

860

Question 5●●○ Standard✍️ By hand · Paper 1 · no calculator[5 marks]

The expressions

k+2

3k-1

and

4k+1

are three consecutive terms of an arithmetic sequence.

(a) Find the value of $k$ .[3]
(b) Hence find the common difference.[2]

Show the full step-by-step solution

Step 1🧠 ThinkThe key fact

In an arithmetic sequence the gap from the 1st to the 2nd term equals the gap from the 2nd to the 3rd. So (middle − first) = (last − middle).

Part (a)

Step 1✍️ By hand

(3k-1) - (k+2) = (4k+1) - (3k-1)

2k - 3 = k + 2

k = 5

Part (b)

Step 1✍️ By hand

Terms:

7,\ 14,\ 21

d = 7

Final answer (a)

k = 5

(b)

d = 7

⚠️ Common mistake Setting the three expressions equal to each other. They are different terms; it is the differences that are equal.

Question 6●●○ Standard✍️ By hand · Paper 1 · no calculator[6 marks]

The sum of the first

n

terms of an arithmetic sequence is

S_n = 3n^{2} + 2n

(a) Find $u_1$ and $u_2$ .[3]
(b) Show that $u_n = 6n - 1$ .[3]

Show the full step-by-step solution

Step 1🧠 ThinkA sum of one term is just that term

S_1

is the sum of the first 1 term, so

S_1 = u_1

S_2 = u_1 + u_2

, so

u_2 = S_2 - S_1

. In general,

u_n = S_n - S_{n-1}

Part (a)

Step 1✍️ By hand

u_1 = S_1 = 3 + 2 = 5

S_2 = 12 + 4 = 16

u_2 = 16 - 5 = 11

Part (b)

Step 1✍️ By handMethod: d, then the nth-term formula

d = 11 - 5 = 6

u_n = 5 + (n-1)\times6

= 5 + 6n - 6 = 6n - 1

💡 Remember Another way:

u_n = S_n - S_{n-1} = 3n^{2}+2n - \left(3(n-1)^{2} + 2(n-1)\right)

. Expanding gives

6n - 1

as well.

Final answer (a)

u_1 = 5,\ u_2 = 11

(b) shown

Question 7●●○ Standard🖩 GDC · Paper 2[4 marks]

An arithmetic sequence has first term

12

and common difference

2.5

. Find the least number of terms for which the sum is greater than

1000

Show the full step-by-step solution

Step 1🧠 ThinkSet up the equation for the sum

Solve

S_n = 1000

with the GDC. The answer will not be a whole number, so we check the whole numbers either side.

Step 2✍️ By hand

\dfrac{n}{2}\left(2\times12 + (n-1)\times2.5\right) = 1000

Step 3🖩 On the GDCnSolve, then test n = 24 and n = 25

menu💻 click menu on keypad›3: Algebra›1: Numerical Solve›n/2·(24+(n−1)·2.5)=1000,n,1,100›enter💻 Enter

The bounds 1, 100 make the GDC look only for a positive

n

S₂₄ = 978 < 1000 < S₂₅ = 1050

Step 4✍️ By hand

n = 24.3\ldots

S_{24} = 978,\ S_{25} = 1050

n = 25

Final answer

25

terms

⚠️ Common mistake Rounding

24.3

down to 24: then the sum is only 978.

Question 8●●○ Standard✍️ + 🖩 · Paper 2[6 marks]

A new theatre in Gaborone has 18 seats in the front row. Each row has 3 more seats than the row in front of it.

(a) Find the number of seats in the 30th row.[2]
(b) Find the total number of seats in the first 30 rows.[2]
(c) The owners need at least 1500 seats. Find the least number of rows needed.[2]

Show the full step-by-step solution

Step 1🧠 ThinkSpot the sequence

Seats per row: 18, 21, 24, … so

u_1 = 18,\ d = 3

. One row is one term; all seats together is a sum.

Part (a)

Step 1✍️ By hand

u_{30} = 18 + 29\times3

= 105

seatsA1

Part (b)

Step 1✍️ By hand

S_{30} = \dfrac{30}{2}\left(18 + 105\right)

= 1845

seatsA1

Part (c)

Step 1🖩 On the GDCSolve S_n = 1500 and check either side

menu💻 click menu on keypad›3: Algebra›1: Numerical Solve›n/2·(36+(n−1)·3)=1500,n,1,100›enter💻 Enter

26 rows: 1443 seats; 27 rows: 1539 seats

Step 2✍️ By hand

n = 26.6\ldots

27

rowsA1

Final answer (a) 105 (b) 1845 (c) 27 rows

Question 9●●● Exam-level✍️ + 🖩 · Paper 2[7 marks]

Mpho saves P 150 in the first month. Each month after that she saves P 20 more than in the month before.

(a) Find how much she saves in the 12th month.[2]
(b) Find her total savings after 12 months.[2]
(c) Find the number of months it takes for her total savings to first reach P 15 000.[3]

Show the full step-by-step solution

Step 1🧠 ThinkMonthly amount = term, total = sum

Amounts saved: 150, 170, 190, … so

u_1 = 150,\ d = 20

Part (a)

Step 1✍️ By hand

u_{12} = 150 + 11\times20

= \text{P }370

Part (b)

Step 1✍️ By hand

S_{12} = \dfrac{12}{2}\left(150 + 370\right)

= \text{P }3120

Part (c)

Step 1🖩 On the GDCSolve S_n = 15 000

menu💻 click menu on keypad›3: Algebra›1: Numerical Solve›n/2·(300+(n−1)·20)=15000,n,1,100›enter💻 Enter

After 32 months: P 14 720; after 33: P 15 510

Step 2✍️ By hand

\dfrac{n}{2}\left(300 + 20(n-1)\right) = 15\,000

n = 32.4\ldots

33

monthsA1

Final answer (a) P 370 (b) P 3120 (c) 33 months

Question 10●●● Exam-level✍️ By hand · Paper 1 · no calculator[6 marks]

Consider the multiples of 7 that lie between 100 and 500.

(a) Write down the smallest and the largest of these multiples.[2]
(b) Show that there are 57 of these multiples.[2]
(c) Find their sum.[2]

Show the full step-by-step solution

Step 1🧠 ThinkMultiples of 7 form an arithmetic sequence

Multiples of 7 go up by 7 each time, so

d = 7

. We need the first one above 100 and the last one below 500.

Part (a)

Step 1✍️ By hand

7\times14 = 98

(too small),

7\times15 = 105

Smallest:

105

7\times71 = 497

7\times72 = 504

(too big)

Largest:

497

Part (b)

Step 1✍️ By hand

105 + (n-1)\times7 = 497

(n-1)\times7 = 392 \Rightarrow n - 1 = 56

n = 57

Part (c)

Step 1✍️ By hand

S_{57} = \dfrac{57}{2}\left(105 + 497\right)

= 57\times301

= 17\,157

💡 Remember

57\times301 = 57\times300 + 57 = 17\,100 + 57

. Splitting numbers like this makes Paper 1 arithmetic safe.

Final answer (a) 105 and 497 (b) shown (c)

17\,157

Question 11●●● Exam-level✍️ + 🖩 · Paper 2[7 marks]

For an arithmetic sequence, the sum of the first 10 terms is 240 and the 15th term is 62.

(a) Write down two equations in $u_1$ and $d$ .[2]
(b) Find $u_1$ and $d$ .[2]
(c) Find the least value of $n$ for which $S_n \gt 2000$ .[3]

Show the full step-by-step solution

Part (a)

Step 1✍️ By handOne equation from the sum, one from the term

\dfrac{10}{2}\left(2u_1 + 9d\right) = 240 \Rightarrow 2u_1 + 9d = 48

u_1 + 14d = 62

Part (b)

Step 1🖩 On the GDCSolve the pair of equations

menu💻 click menu on keypad›3: Algebra›2: Solve System of Linear Equations…

Choose 2 equations, variables a, d (a stands for

u_1

). Type each equation and press enter.

u₁ = 6, d = 4

Step 2✍️ By hand

u_1 = 6,\ d = 4

A1A1

Part (c)

Step 1🖩 On the GDCSolve S_n = 2000 and check

menu💻 click menu on keypad›3: Algebra›1: Numerical Solve›n/2·(12+4·(n−1))=2000,n,1,100›enter💻 Enter

S₃₀ = 1920, S₃₁ = 2046

Step 2✍️ By hand

\dfrac{n}{2}\left(12 + 4(n-1)\right) \gt 2000

n \gt 30.6\ldots

n = 31

Final answer (a)

2u_1 + 9d = 48

u_1 + 14d = 62

(b)

u_1 = 6,\ d = 4

(c)

n = 31

Question 12●●● Exam-level✍️ By hand · Paper 1 · no calculator[5 marks]

Show that

\displaystyle\sum_{r=1}^{n}(2r+3) = n(n+4)

. Hence find the value of

n

for which

\displaystyle\sum_{r=1}^{n}(2r+3) = 192

Show the full step-by-step solution

Step 1🧠 ThinkIt is an arithmetic series with a letter for the number of terms

The terms are

5, 7, 9, \ldots

u_1 = 5,\ d = 2

, and there are

n

terms. Use the sum formula, then simplify until it matches

n(n+4)

Step 2✍️ By handShow that

S_n = \dfrac{n}{2}\left(2\times5 + (n-1)\times2\right)

= \dfrac{n}{2}\left(2n + 8\right)

= n(n+4)

Step 3✍️ By handHence: solve the quadratic

n^{2} + 4n - 192 = 0

(n + 16)(n - 12) = 0

n = 12

(since

n \gt 0

)A1

✅ Check it

12\times16 = 192

✔

Final answer Shown;

n = 12

⚠️ Common mistake Keeping

n = -16

. The number of terms cannot be negative.

🆘 Stuck on a question? Open the notes, GDC toolkit and common mistakes

💻 Using the TI-Nspire CX Student Software on a laptop

The screens in these notes are drawn exactly as the Student Software shows them. The keypad on the left of the software works like the handheld — click its keys with the mouse.
menu: click menu on the on-screen keypad (or Documents Toolbox ▸ 🔧). The menu numbers and letters are the same as in these notes.
enter = Enter · tab = Tab (next box) · esc = Esc · del = Backspace.
Powers: type ^ (Shift+6), the power, then press → to come back down before typing more. x² = ^2 then →.
Fraction: type / · √: type sqrt( · π: type pi · eˣ: type exp( · store: type :=. For EE (×10ⁿ) and i, click them on the keypad.
New page: Ctrl+I (or the Insert menu) · change page: Ctrl+← / Ctrl+→ · graph entry line: Ctrl+G · var: Ctrl+L.
Radian / Degree: click Settings in the status bar at the bottom of the window. The mode (RAD or DEG) shows in the top bar.
IB exam mode (Press-to-Test): everything in these notes works on the non-CAS TI-Nspire with the IB exam settings. In exam mode there are no dot-product, cross-product, unit-vector or eigenvalue keys and no Diff Eq graphs — the notes show the allowed way (type the components, use polyRoots on the characteristic equation, do Euler steps on a Calculator page).
If you type nDeriv(, the software shows it as centralDiff( — it is the same command.

🎯 What you must know

Arithmetic sequence: the same number $d$ (the common difference) is added each time. $d = u_2 - u_1$ ; $d$ is negative if the terms go down.
$n$ th term: $u_n = u_1 + (n-1)d$ . From term 1 to term $n$ there are $n-1$ jumps.
Sum of the first $n$ terms: $S_n = \dfrac{n}{2}\left(2u_1+(n-1)d\right) = \dfrac{n}{2}\left(u_1+u_n\right)$ .
Sigma notation: $\displaystyle\sum_{k=1}^{20}(3k+2)$ means put $k=1,2,\ldots,20$ into $3k+2$ and add. The number of terms is top − bottom + 1.
Three terms $a, b, c$ are consecutive terms of an arithmetic sequence exactly when $b - a = c - b$ , i.e. $2b = a + c$ .
If you know $S_n$ as a formula, then $u_1 = S_1$ and $u_n = S_n - S_{n-1}$ .
$n$ must be a positive whole number. If an equation gives $n = 24.3$ , check the whole numbers either side.

nth term

u_n = u_1 + (n-1)d

📘 In the formula booklet

Sum of n terms

S_n = \dfrac{n}{2}\left(2u_1+(n-1)d\right) = \dfrac{n}{2}\left(u_1+u_n\right)

📘 In the formula booklet

🖩 Your GDC toolkit for this subtopic

Follow the steps on your own TI-Nspire CX II. The screens show what you should see.

List terms and add them: sum(seq(formula, n, first, last))

sum(seq(5+4·(n−1),n,1,25))›enter💻 Enter

📍 Where to find it on the TI-Nspire

sum(is inmenu💻 click menu on keypad›6: Statistics›3: List Math›5: Sum of Elements

seq(is inmenu💻 click menu on keypad›6: Statistics›4: List Operations›5: Sequence

Type sum(seq(5+4·(n−1),n,1,25)).
Press enter.

S₂₄ = 978 < 1000 < S₂₅ = 1050

Worked example: Question 7 — nSolve, then test n = 24 and n = 25

The bounds 1, 100 make the GDC look only for a positive

n

Solve an equation for n (inserts nSolve( ); add bounds 1, 100 to get the positive answer)

menu💻 click menu on keypad›3: Algebra›1: Numerical Solve

Start on a Calculator page (ctrl doc adds one).
Press menu.
Choose 3: Algebra.
Choose 1: Numerical Solve.

The menu path

S₂₄ = 978 < 1000 < S₂₅ = 1050

Worked example: Question 7 — nSolve, then test n = 24 and n = 25

The bounds 1, 100 make the GDC look only for a positive

n

Find u₁ and d from two equations

menu💻 click menu on keypad›3: Algebra›2: Solve System of Linear Equations…

Start on a Calculator page (ctrl doc adds one).
Press menu.
Choose 3: Algebra.
Choose 2: Solve System of Linear Equations….
A box opens. Fill in each field (press tab to move between fields), then press enter. The results appear on the screen.

The menu path

u₁ = 6, d = 4

Worked example: Question 11 — Solve the pair of equations

Choose 2 equations, variables a, d (a stands for

u_1

). Type each equation and press enter.

Sigma template

templates💻 click ⊞ on keypad›Σ

📍 Where to find it on the TI-Nspire

Σ templateis inmenu💻 click menu on keypad›4: Calculus›3: Sum

Press the templates key and pick the template.
Choose Σ.

⚠️ Where students lose marks

Using $u_1 + nd$ instead of $u_1 + (n-1)d$ .
Mixing up $u_n$ (one term) and $S_n$ (the total).
Counting the terms in $\displaystyle\sum_{k=5}^{20}$ as 15. It is $20-5+1 = 16$ .
Forgetting brackets: $2u_1+(n-1)d$ — the $(n-1)$ multiplies only $d$ .
Giving $n = 24.3$ as the answer, or rounding the wrong way. Test $S_{24}$ and $S_{25}$ .

IB DP Mathematics: Analysis and Approaches (AA) — SL and HL AA 1.2 — Arithmetic sequences and series

IB DP Mathematics: Analysis and Approaches (AA) — SL and HL

1.2 Arithmetic sequences and series

✏️ Practice questions

🎯 What you must know

🖩 Your GDC toolkit for this subtopic

⚠️ Where students lose marks