Arithmetic sequences and their sums in the Mathematik (CBK) exam at WU

The first of the nine topics in the syllabus of Mathematik in the CBK at WU Vienna is called "Lineare und quadratische Funktionen" (linear and quadratic functions); the syllabus does not list its subtopics.¹ The textbook by Birgit Rudloff, who leads the course, and Achim Zeileis has a first chapter with the same title, and arithmetic sequences are in it: lists of numbers that go up or down by the same amount at every step, such as output that grows by the same number of units each month, or instalments that shrink by the same amount each quarter. The task is then to find a single value, the total over several periods, or the number of periods until a total reaches a target.

You need two formulas for that, one for a single term and one for the sum, and for the third calculation you solve a quadratic equation. Take most care with counting: miscount by one period and you are exactly one step off, and with five options that number may well be among them. German terms are given in brackets.

The n-th term of an arithmetic sequence: Mathematik (CBK)

An arithmetic sequence is fixed by two numbers: the first term $a_1$ and the common difference $d$ (Differenz), the amount by which each term exceeds the one before. If the values fall, $d$ is negative; if they stay level, it is zero. By the $n$-th term the difference has been added $n-1$ times, because the first term has none yet:

$$a_n=a_1+(n-1)\cdot d$$

An example: a bike rental hires out 20 bikes on the first day of the season and, after that, 5 more each day than the day before. On day 14 that is $a_{14}=20+13\cdot 5=85$ bikes. Working with $14\cdot 5$ instead of $13\cdot 5$ gives 90, one day too far.

As a function of $n$ the formula is linear, with slope $d$, which ties sequences to linear functions. That view helps when you are given two arbitrary terms instead of $a_1$ and $d$: $d$ is the change in value divided by the gap between the positions, $d=(a_m-a_k)/(m-k)$. For the bike rental, from $a_4=35$ and $a_{10}=65$: days 4 and 10 are 6 steps apart, so $d=30/6=5$, and going back 3 steps to day 1 gives $a_1=35-3\cdot 5=20$.

The sum of an arithmetic sequence: Mathematik (CBK)

To add up the first $n$ terms you do not need each one. The first and last terms add up to the same as the second and second-to-last, since one is $d$ larger and the other $d$ smaller. So the sum is the number of terms times the mean of the first and last term:

$$\begin{array}{rl}{s}_n& =n\cdot \frac{a_1+a_n}{2}\\ & =\frac{n}{2}\cdot (2a_1+(n-1)\cdot d)\end{array}$$

Over the first 14 days the rental hires out $s_{14}=14\cdot (20+85)/2=735$ bikes. The second form is for when $a_n$ is not given; it already has the $n$-th term built in.

14 · 105 / 2 = 735 bikes

Our calculation

Numbers from the article’s example (our own numbers, not an exam question): a₁ = 20, d = 5, so a₁₄ = 20 + 13 · 5 = 85.

The reversed sequence starts at 85 and falls by 5. On every day k, a_k and a_15−k sit on top of each other, and they change by 5 in opposite directions, so every column is equally high: 20 + 85 = 25 + 80 = … = 85 + 20 = 105.

The two staircases fill a rectangle of 14 · 105 = 1470. One staircase is half of it: 1470 / 2 = 735. Added one by one: 20 + 25 + 30 + 35 + 40 + 45 + 50 + 55 + 60 + 65 + 70 + 75 + 80 + 85 = 735.

The first form needs the $n$-th term, $a_1+(n-1)\cdot d$. Putting in $a_1+n\cdot d$ uses the term after it as the last term. Counting the terms is the other place to slip: from day 5 to day 14 there are $14-5+1=10$ days, because both ends count.

Periods until a target: Mathematik (CBK)

The third calculation turns the sum around and asks after how many periods the total reaches a given value. Now $n$ is the unknown, and the sum formula becomes a quadratic equation. Say you want the day after which the rental has hired out at least 600 bikes in total. The total after $n$ days is $\frac{n}{2}\cdot (40+(n-1)\cdot 5)=\frac{n}{2}\cdot (35+5n)$, and multiplying by $\frac{2}{5}$ turns the condition into:

$$\begin{array}{rl}\frac{n}{2}\cdot (35+5n)& =600\\ n\cdot (7+n)& =240\\ n^2+7n-240& =0\end{array}$$

The quadratic formula gives the positive solution $n=(-7+\sqrt{7^2+4\cdot 240})/2=-3.5+\sqrt{252.25}\approx 12.4$. The negative one means nothing here, since a number of days cannot be negative. After 12 days the total is only $s_{12}=570$ bikes, after 13 days $s_{13}=650$. The answer is 13: the question asks for the first whole period after which the target is reached, so you round up, however small the decimal part.

n ≈ 12.4, rounded up 13 days

Our calculation

Numbers from the article’s example (our own numbers, not an exam question): a₁ = 20, d = 5, target 600 bikes.

Total after n days: s_n = n/2 · (40 + (n − 1) · 5) = n/2 · (35 + 5n). Each column is the one before plus the new day’s bikes.

day	1	2	3	4	5	6	7	8	9	10	11	12	13
bikes that day	20	25	30	35	40	45	50	55	60	65	70	75	80
in total	20	45	75	110	150	195	245	300	360	425	495	570	650

The curve is the same formula for every n, between two days as well. It reaches 600 where n/2 · (35 + 5n) = 600, that is n² + 7n − 240 = 0, positive solution n = −3.5 + √252.25 ≈ 12.38. There is no day 12.4: after day 12 the total is 570, after day 13 it is 650, so 13.

Those two final sums double as the check for any answer of this kind: with the rounded number the total must reach the target, with one period fewer it must fall short. Rounding down, or using the term instead of the sum, fails one of the two.

Practice questions on arithmetic sequences: Mathematik (CBK)

The exam has 20 multiple-choice questions with 5 options each, exactly one of them right. A right answer earns 1 point. Your first 3 wrong answers cost nothing, every further one costs 0.25 points, and you pass with 11 points.¹

We wrote the three questions for this article in that format. The stories, numbers and wording are our own; none of them comes from a WU exam or from the textbook. Every wrong option is the result of a typical mistake. Work each one out first; the solution then shows the working, the typical mistake and how to rule options out.

Practice question 1Linear and quadratic functions

A start-up wins 140 new customers in its first month and, after that, 18 more each month than in the month before. How many new customers does it win in month 15?

1. a)252
2. b)266
3. c)270
4. d)392
5. e)410

Exactly one option is right.

Solution and points for every choice+

Worked solution

1. First term a₁ = 140, common difference d = 18.
2. By month 15 the increase has been added 14 times: a₁₅ = a₁ + (15 − 1) · d.
3. a₁₅ = 140 + 14 · 18 = 392.

a)

Wrong. Only the increases counted, the starting value missing: 14 · 18.

0 or −0.25

b)

Wrong. The average of the first 15 months instead of the value in month 15: (140 + 392) / 2.

0 or −0.25

c)

Wrong. Starting value forgotten and 15 instead of 14 increases: 15 · 18.

0 or −0.25

d)

Right. a₁₅ = 140 + (15 − 1) · 18 = 392.

+1

e)

Wrong. n instead of n − 1: 140 + 15 · 18.

0 or −0.25

–

Left blank.

0

Typical mistake. n instead of n − 1. The first month has no increase yet, which is why the formula has n − 1. With 15 increases you get 410.

Ruling out before you finish. Any possible answer is 140 plus a multiple of 18. 252 and 270 are 112 and 130 above 140, neither a multiple of 18. That rules out options a and c and leaves three.

A guess among the 3 left has a 1-in-3 chance. It is worth ≈ 0.33 points on average while your three free wrong answers last, and ≈ 0.17 after that.

Practice question 2Linear and quadratic functions

A company repays a loan in quarterly instalments that fall by the same amount every quarter. The 3rd instalment is 880 MU (monetary units), the 8th is 780 MU. How much does it pay in total with the first 12 instalments?

1. a)8,760 MU
2. b)8,910 MU
3. c)9,240 MU
4. d)9,600 MU
5. e)9,720 MU

Exactly one option is right.

Solution and points for every choice+

Worked solution

1. From the 3rd to the 8th instalment there are 8 − 3 = 5 steps: d = (780 − 880) / 5 = −20.
2. First instalment a₁ = a₃ − 2d = 880 + 40 = 920, twelfth a₁₂ = 920 + 11 · (−20) = 700.
3. s₁₂ = 12 · (a₁ + a₁₂) / 2 = 12 · (920 + 700) / 2 = 9,720 MU.

a)

Wrong. The sign flipped when working back to the 1st instalment: a₁ = 880 − 40 = 840, a₁₂ = 620.

0 or −0.25

b)

Wrong. Multiplied by 11 instead of 12 instalments: 11 · (920 + 700) / 2.

0 or −0.25

c)

Wrong. The 3rd instalment taken as the first: 12 · (880 + 660) / 2.

0 or −0.25

d)

Wrong. The wrong last term: a₁ + 12d = 680 instead of a₁₂ = a₁ + 11d = 700.

0 or −0.25

e)

Right. s₁₂ = 12 · (920 + 700) / 2 = 9,720.

+1

–

Left blank.

0

Typical mistake. The wrong last term. The sum formula needs the 12th instalment, a₁ + 11d. Using a₁ + 12d = 680 means working with the 13th instalment, and gives 9,600 MU.

Ruling out before you finish. The instalments fall. The 1st and 12th, the 2nd and 11th and so on each add up to the same as the 6th and 7th, so the average of all 12 is the average of the 6th and 7th. It lies below the average of the 3rd and 8th, 830 MU, and above the 8th, 780 MU. So the total lies between 12 · 780 = 9,360 and 12 · 830 = 9,960 MU. That rules out options a, b and c and leaves two.

A guess among the 2 left has a 1-in-2 chance. It is worth 0.5 points on average while your three free wrong answers last, and 0.375 after that.

Practice question 3Linear and quadratic functions

An online shop ships 60 parcels in its first week and, after that, 8 more each week than the week before. After how many weeks has it shipped at least 1,000 parcels in total?

1. a)10
2. b)11
3. c)16
4. d)17
5. e)119

Exactly one option is right.

Solution and points for every choice+

Worked solution

1. Total after n weeks: sₙ = n · (2 · 60 + (n − 1) · 8) / 2 = 4n² + 56n.
2. 4n² + 56n = 1,000 is equivalent to n² + 14n − 250 = 0; the quadratic formula gives the positive solution n = (−14 + √(14² + 4 · 250)) / 2 = −7 + √299 ≈ 10.3.
3. After 10 weeks it is 960 parcels, after 11 weeks 1,100. The answer is 11.

a)

Wrong. Correct calculation, rounded down: after 10 weeks it is only 960 parcels.

0 or −0.25

b)

Right. n ≈ 10.3, rounded up: after 11 weeks it is 1,100 parcels.

+1

c)

Wrong. Calculated without the increase and rounded down: 1,000 / 60 ≈ 16.7.

0 or −0.25

d)

Wrong. Calculated without the increase: 1,000 / 60 ≈ 16.7, rounded up.

0 or −0.25

e)

Wrong. The term instead of the sum: only in week 119 does the shop ship at least 1,000 parcels in a single week.

0 or −0.25

–

Left blank.

0

Typical mistake. Rounding down. n ≈ 10.3 means the threshold has not been reached after 10 weeks. The question asks for the first whole week after which the total is at least 1,000, so round up.

Ruling out before you finish. At a flat 60 parcels a week, 16 weeks would already give 16 · 60 = 960, and the increases of weeks 2 to 4 alone add 8 + 16 + 24 = 48. So after 16 weeks the total is above 1,000 and the answer is at most 16: that rules out options d and e and leaves three.

A guess among the 3 left has a 1-in-3 chance. It is worth ≈ 0.33 points on average while your three free wrong answers last, and ≈ 0.17 after that.

Exam rules and permitted aids for Mathematik (CBK)

WU does not publish which of the 20 questions covers which chapter. This article follows the structure of the textbook, where arithmetic sequences and their sums (Arithmetische Folgen, Summierung von Folgen) are a section of the first chapter.²

You may use a calculator, but not one with functions for differential calculus, integration or matrices, not one that solves linear systems, and not one with a text memory. If German is not your first language you may bring a dictionary; the syllabus allows nothing else.¹ For the quadratic equation, use the quadratic formula and take the square root on the calculator.

WU's guideline for conducting exams bans formula sheets and watches of any kind unless the syllabus says otherwise.³ So you cannot bring your own formula sheet, and WU does not say whether one is handed out. Know the formulas for $a_n$ and $s_n$ by heart. If the sum formula slips your mind, pairing the first and last terms rebuilds it.

Next step after arithmetic sequences: Mathematik (CBK)

1. Read the section on arithmetic sequences and their sums in the free textbook by Birgit Rudloff and Achim Zeileis.⁴ Then do the three questions above without looking at the solutions first.
2. If you got one wrong, find the mistake in the solution that matches your answer. If it was a counting slip, write down before calculating which term is asked for and how many terms the sum has.
3. Cost, revenue and the monopoly from the same chapter are covered in our article on cost, revenue and monopoly. Exam dates and registration are in the overview of the Mathematik (CBK) exam.

Sources

1. 1Syllabus 0003 Mathematik (LVP), Wintersemester 2026/27WU Wien, Vorlesungsverzeichnis · undated page · checked on 27 September 2026
2. 2Rudloff, Zeileis: Mathematik für Wirtschaftswissenschaften, Kapitel 1: Lineare und quadratische Funktionenmathe4wiwi.org · undated page · checked on 26 September 2026
3. 3Richtlinie zur Abhaltung von Präsenzprüfungen, 2026 (PDF)WU Wien · document dated 2 March 2026 · checked on 27 September 2026
4. 4Rudloff, Zeileis: Mathematik für Wirtschaftswissenschaften (Online-Buch, Version 2023)mathe4wiwi.org · undated page · checked on 26 September 2026