Written Guide

A relatively challenging component of Numerical Reasoning is the Number Series Questions. These questions require the candidate to identify the pattern between a series of numbers in order to discern the next number in the sequence.

These sequences can vary in complexity, and require different approaches to identification. In the test, it is absolutely vital to approach questions with some urgency as time is extremely limited. One should take care not to linger too long on any single question as it takes up valuable time which would be better-allocated on other, more approachable questions.

While this guide will explore the means by which to approach the number series questions, it is vital that if you cannot identify a path to a solution within 10-12 seconds you take an educated guess and move on to the next question. Spending too much time on any single question can cost precious time which is better spent elsewhere on less challenging problems.

In order to understand the more complex series types, let’s first examine the two basic types: Linear Series and Geometric Series.



Linear Series

Linear series are number sequences in which the sequence progresses in a linear fashion. This means that the amount by which the number changes to the next step is constant. For a sequence to be linear, it must feature either addition or subtraction:

Example: Addition

Question: What is the next number in the series?



-1 , 5 , 11 , 17 , 23 , _?_



• 28
• 29
• 22
• 24

Answer: 29. The number increases by 6 with each step.



-1 , 5 , 11 , 17 , 23 , 29

⤻ ⤻ ⤻ ⤻ ⤻

+6 +6 +6 +6 +6



Example: Subtraction

Question: What is the next number in the series?

72 , 60, 48, 36, 24, _?_



• 10
• 18
• 12
• 14

Answer: 12. The number decreases by 12 with each step.

72 , 60 , 48 , 36 , 24 , _?_

⤻ ⤻ ⤻ ⤻ ⤻

-12 -12 -12 -12 -12



These questions are certainly easy to solve. Unfortunately, however, they are very uncommon in the test as almost anybody could solve them with ease – after all, what use is an intelligence test that everybody passes easily? Far more commonly featured in the test are the following sequence types.

Geometric Series

Geometric series, also referred to as a geometric progression, are series where the next number is found by multiplying or dividing the previous number by a fixed number. These sequences are not linear, as the number will increase or decrease by a higher amount with each step.

Example: Multiplication

Question: What is the next number in the series?

 

7 , 21 , 63 , 189 , 567

⤻ ⤻ ⤻ ⤻

x3 x3 x3 x3



• 1534
• 1701
• 1890
• 2042

Answer: 1701. The number is multiplied by 3 with each step.



Example: Division

Question: What is the next number in the series?



15872 , 3968 , 992 , 248 , 62

⤻ ⤻ ⤻ ⤻

/4 /4 /4 /4



• 31
• 15.5
• 7.5
• 16

Answer: 15.5. The number is divided by 4 with each step.



As you can see, the calculation required for geometric series questions can be considerably harder than that required for linear series questions. However, using the tips provided you can easily identify these relationships early on.

How to Solve a Geometric Series

When we have decided that a series is geometric, we must then identify whether it is using division or multiplication. This is quite straightforward. Ignoring negative values (as we cannot multiply/divide a positive/negative number into the other, we can treat them all as positive numbers in a sequence – just don’t forget to include the negative notation in your answer if the sequence is negative), if the number is increasing then it is being multiplied, and if it is decreasing it is being divided. For example:

Question: What is the next number in the series?

4 , 12 , 36 , 108 , 324

• 648
• 972
• 822
• 784

In this question, we need to figure out by how much the sequence is changing. As it is increasing in seemingly consistent increments, we can be certain that this is a multiplication series.

In order to determine the amount by which these numbers are being multiplied, we should take a look at the first two numbers in order to understand the pattern of the series.

As the first two numbers in the sequence will be the lowest in the instance of a multiplication series, it is easier to calculate the multiplication value this way.

To this end, let’s start by dividing the second number by the first:

12/4=3

So we have the hypothesis that the multiplier is 3. Let’s verify this by multiplying the second number by this amount:

12×3=36

This can be followed through the rest of the numbers, but it is usually sufficient to simply apply the multiplier to the final number. Here is each step of the multiplier.



4 , 12 , 36 , 108 , 324

⤻ ⤻ ⤻ ⤻

+6 +6 +6 +6



It would seem that we have the correct multiplier. Now, let’s multiply the final number to reach our answer:

324×3=972

There we have it – the answer is 972.



In the case of division series, the approach is quite similar. We simply take the approach the other way round. Here’s an example:

Question: What is the next number in the series?



1024 , 256 , 64 , 16 , 4

• 2
• 1
• 2.5
• 3

Now, as this is a division series, we can treat it in reverse as if it is a multiplication series from 4 to 1024. Now, in order to find the divisor, we can divide the second-to-last number by the final number::

16/4=4

So, it would seem that 4 is the divisor. Lets verify:

16×4=64

Now, to use this, we do not multiply the highest number as we did with the multiplication series – instead, we divide the final number by our divisor:

4/4=1

By following this methodology, we have reached an answer of 1.



Now that we understand the two foundational series types, we can use this knowledge to approach the more complex series types: Compound Series and Layered Series. These two series are quite a bit more complicated than what we have explored so far, but with the right practice and understanding of what to expect, you can be prepared for any of the sequences given in the test.



Compound Series

A compound series is a sequence of numbers in which the value changes based on alternating, unrelated rules. In simpler cases, this could be as follows:

12 , 18 , 14 , 20 , 16

⤻ ⤻ ⤻ ⤻

+6 -4 +6 -4

As we can see, the amount by which the number changes is doing so in an alternating manner – it increases by six, then decreases by 4. In this instance, the next number in the sequence would be 22.

However, these sequences can be considerably more complicated:



34 , 17 , 102 , 51 , 306

⤻ ⤻ ⤻ ⤻

/2 x6 /2 x6

In this instance, the sequence alternates between multiplication and division. In this case, the next operation to implement is division by 2, giving an answer of 153.

In these compound sequences, the alternating pattern can feature any of the four basic operations: addition, subtraction, multiplication and division, in any combination.

How to Solve a Compound Series

When it has been identified that a sequence constitutes a compound series, the approach to solving this is quite simple. First, we must identify the relationship between two pairs of numbers: the first and second, and the second and third. For example:

Question: What is the next number in the series?



4 , 12 , 6 , 18 , 12

• 24
• 36
• 18
• 16



Now, as we can see, the numbers appear to be following two separate sets of rules. Let’s examine this pattern and find out how to effectively reach a conclusion within an effective time frame.

As we stated earlier, we begin by taking a look at the two pairs of numbers – the first and second, and the second and third, for a relationship. Here, we essentially want to be eliminating the options for transition which are not consistent across the series. To begin with, we have 4 and 12. Now, this transition can be reached by two simple means:

4×3=12

and

4+8=12

As there is no single pattern between these two, we must investigate further.

Next, let’s take a look at the second and third numbers in the sequence: 12 and 6. Again, these answers can be reached by two simple means:

12/2=6

and

12-6=6

Once again, no single pattern is evident. Therefore, we must check another two pairs to verify which of the two options is the relation between the first two pairs. In this case, it is the third and fourth and the fourth and fifth numbers which constitute the pairs for comparison.

The third and fourth, 6 and 18, can be reached by two means:

6×3=18

and

6+12=18

It would seem that multiplying by 3 is the correct approach, as it is consistent in both pairings.Therefore, we can assume that the first of the two alternating rules is to multiply by 3.

The fourth and fifth, 18 and 12, can be reached by just one simple means:

18-6=12

As there is only one simple solution, and it is consistent with the second pairing, it is safe to assume that the second of the alternating rules is to divide by 6. As such, we reach the following model:



4 , 12 , 6 , 18 , 12

⤻ ⤻ ⤻ ⤻

x3 -6 x3 -6

As such, we can see that the next step to be taken is to multiply by 3:

12×3=36

Therefore, the answer is 36.

In these questions, the key is that you identify the nature of the series quickly. In this case, when you see a series that goes up, then down, then up again, and so on, you should suspect that it is a compound series. In the first pair, the transition is either x3 or +8; in the next pair it is either /2 or -6; in the third pair it is either x3 or +12. At this point, we can be confident that the first transition of the compound pattern is x3 – all that remains then is to determine from the last pair whether the second transition is /2 or -6. In the case of the example we discussed, the shape is very clearly seen, almost as if scaling a set of stairs:



Make sure to check for a constant value for the transition between numbers; x and y as shown below:

a , b , c , d , e

⤻ ⤻ ⤻ ⤻

x y x y



Layered Series

A layered series features a sequence within a sequence – the amounts by which the main sequence changes constitute a sequence of their own, following an independent rule. With this type of question, the key to recognition lies in the inequality between gaps. It is easy to see in these sequences that the ratio between the numbers is not constant – numbers are not being added to, subtracted from, divided by or multiplied by any common number, and it is plain to see in the increments of change.

For questions such as this, it is highly recommended to use scrap paper to write down and track the numbers within the sequence and subsequent relationships between those numbers. By doing so, it is far easier to visually track what is going on, and it will often be the case that the pattern of the gaps within the sequence will stand out visually.

For example:



27 , 34 , 48 , 76 , 132

⤻ ⤻ ⤻ ⤻

+7 +14 +28 +56

⤻ ⤻ ⤻

x2 x2 x2

Here, we see that the numbers by which the sequence changes are changed by their own set of rules. In this instance, the number being changed by is doubled with each step. As such, the next increase is 112, making the next step in the sequence 244. This case is not too complicated as the entire sequence is consistently increasing. In some cases, this might not be the case.

 

3 , 20 , 21 , 6 , -25

⤻ ⤻ ⤻ ⤻

+17 +1 -15 -31

⤻ ⤻ ⤻

-16 -16 -16



As we can see in this example, the number sequence increases to a peak then begins to decrease again. This is because the second layer of the sequence is subtracting 16 from the amount by which the sequence is changing. Although the beginning change had a positive value, the second sequence quickly reduced it to a negative value.

With layered series, it is common to notice a certain curvature in the sequence; in this case, the sequence increases to a peak of 21 then begins to drop again, on account of the second layer of the sequence. If you notice that a series reaches a peak or a trough somewhere in the middle of the sequence and then moves back in the opposite direction, then it is a layered series.



How to Solve a Layered Series

When approaching a layered series, we must take the increments by which the series is changing as a sub-series to be solved before we can identify the pattern of the full series. Let’s have a look at an example:

Question: What is the next number in the series?



24 , 34 , 40 , 42 , 40

• 36
• 34
• 28
• 38

Let’s begin by identifying the jumps in the sequence. It would appear that the sequence is increasing with diminishing returns until it begins to decrease:

24 , 34 , 40 , 42 , 40

⤻ ⤻ ⤻ ⤻

+10 +6 +2 -2



Here, we can see that the relationship is, in fact, addition with diminishing returns until a point of subtraction. Now, to identify the nature of these diminishing returns, we must treat the additions as a sequence of their own. This gives a sequence of +10 • +6 • +2 • -2. Now, as the amount by which the numbers are changing in this sequence is constant, we can deduce that this is a subtraction relationship. Upon inspection, it becomes clear that the amount being subtracted by is 4. With this knowledge, we can untangle this problem.

24 , 34 , 40 , 42 , 40

⤻ ⤻ ⤻ ⤻

+10 +6 +2 -2

⤻ ⤻ ⤻

-4 -4 -4

Now, the problem is as simple as dividing subtracting 4 from -2 to find the next step in the full sequence:

-2 – 4 = -6

So we subtract 6 from the sequence:

40 – 6 = 34

This gives our final answer: 34.



24 , 34 , 40 , 42 , 40 , 34

⤻ ⤻ ⤻ ⤻ ⤻

-12 -12 -12 -12 -12

⤻ ⤻ ⤻ ⤻

-4 -4 -4 -4

Once again, we see a certain curvature in the sequence. In this case, it goes up and then down again. When seeing a curve like this, one can be confident that this is a layered series, though not all layered series are guaranteed to feature such a curve; in some cases, the returns will not diminish to a point that they become negative.

To quickly solve such a question, one should focus on the last three numbers in the sequence in question.

   

A , B , C , D , E

⤻ ⤻ ⤻ ⤻

x y

⤻ ⤻ ⤻

z

As there is a time crunch in the test, it is important to take any shortcuts which will not affect performance negatively. In this case, by focusing on x, y and z we can quickly and effectively solve this type of series. By solving for z, we can then figure out which process to apply to y so that we can find our answer.



Square Number Series

Another type of series which may be encountered is a Square Number Series. In these sequences, a series of square numbers will be given which are squares of a given base number. For instance:



16 , 25 , 36 , 49 , 64

In this instance, each component is a square of a given root number – in this case being 4, 5, 6, 7 and 8. In this instance, the solution is very simple. As a square number is simply the root number multiplied by itself, we multiply 9 by 9 for the next step in the sequence, giving us an answer of 81.

Interestingly, sequences of square numbers can also be treated as a layered series in which the change increases by two at every step. Let’s take a look at the example given to demonstrate.



16 , 25 , 36 , 49 , 64 , 81

⤻ ⤻ ⤻ ⤻ ⤻

+9 +11 +13 +15 +17

⤻ ⤻ ⤻ ⤻

+2 +2 +2 +2

Here we can see that the Square Number sequence is a Layered Series where the second layer is +2 (or in cases where it is going backwards, -2). By knowing this, we can easily find the next step in any Square Number sequence by changing the increment of change by 2, which is considerably quicker than the conventional solution to both a Square Series and a Layered Series.

As such, if you have identified a Square Number sequence, then you may use this shortcut in order to reach a quick solution.

Exponential Series

In certain instances, you may be faced with an Exponential Series. While these are rarer than the other series types, it is still important to understand how to identify and quickly solve these.

An Exponential Series features a sequence of numbers based on a base number which is multiplied by a rising exponent. Simply put, an exponent represents the number of times by which to multiply a number by itself. For instance, ‘two to the power of three’, written as 2^3, means that two needs to be multiplied by itself three times, as such:

2 x 2 x 2 = 8

In the cases of an Exponential Series, we will often see the numbers increase dramatically, or as is said of extreme growth, exponentially. Let’s take a look at an example:



3 , 9 , 27 , 81 , 243

⤻ ⤻ ⤻ ⤻

^2 ^3 ^4 ^5

In this case, we can see that the series is making use of a base number of 3, with an increasing exponent. To solve this, we can simply add one more to the exponent, which means multiplying the latest number in the series by three once more, thus increasing the exponent to ^6:

243 x 3 = 729

As such, the sequence is as follows:

 

3 , 9 , 27 , 81 , 243 , 729

⤻ ⤻ ⤻ ⤻ ⤻

^2 ^3 ^4 ^5 ^6

It is worth noting that an exponential series using a base number is no more, in fact, than a multiplication series which happens to also follow an exponential structure. If we were to multiply each step by 3, we would reach the exact same conclusion:



3 , 9 , 27 , 81 , 243 , 729

⤻ ⤻ ⤻ ⤻ ⤻

x3 x3 x3 x3 x3

To this end, it is usually advisable to treat this manner of question as if it were a multiplication problem – to raise a power by one we need only multiply the latest number in the series by the first number in the series.

How to Identify a Series Type

As we have laid out, there are six types of series seen in the test: Linear Series, Geometric Series, Compound Series, Layered Series, Square Number Series, and Exponential Series. In order to quickly identify the type of series, follow this flowchart:



By following this decision tree and then applying the appropriate method as outlined in this guide, you can ensure that you are fully prepared for this section of the test. If you are not clear on any of the concepts explored here, feel free to read this guide again, or proceed to some practice questions for some hands-on experience. Good luck!