In this post we will talk in depth about the alphabetical series, also known as letter series, and that are widely used in personnel selection processes, oppositions and psychotechnical tests usually. If you prefer, you can also watch this video entry.
We will teach you how to overcome this type of series and we will reveal all its secrets.
We recommend that you review our digital series video since most alphabetic series are nothing more than a specific case of those.
Alphabetical series are presented as a set of letters that follow a logical order that we will have to discover, to deduce the next letter of the series.
To solve these types of questions with ease and minimize errors, it is very important to master the alphabetical order and know the position of each letter in it. Thus, for example, the letter "A" is associated with the number 1, since it occupies the first position of the alphabet, the letter "B", is associated with the number 2 and so on until the letter "Z" occupying the position 27 in the Spanish alphabet. The alphabet must be considered cyclically, that is, after the letter "Z" the "A" would continue and so on.
Normally, the double letters: "CH", "LL" and "RR" are not considered part of the alphabet when solving the series, although whenever possible, it is convenient to ask the examiner.
- 0.1 Simple Alphabetical Series
- 0.2 Multiple interleaved alphabetic series
- 0.3 Mixed series
- 0.4 Alterations and variations
- 0.5 Literal series
- 1 Special cases
Simple Alphabetical Series
These are the simplest series and the ones we are sure to find in any psychotechnical test. Let's put an example:
B D F H?
If we look, we can see that the alphabetical order of the letters increases progressively.
If we substitute each letter for the numerical value corresponding to the position of each one within the alphabet, the previous series becomes this one, which we will call "base series":
2 4 6 8 ?
And if we remember what was learned in the digital series video, we will see that there is an increase of +2 units between every two elements of the base series:
We therefore have an arithmetic series of fixed factor (+2), so the following sequence value will be obtained by adding 2 to the last element of the series, that is: 8 + 2 = 10.
Now we have to look for the letter that occupies the tenth position of the alphabet, which is the "J", and this is the correct answer.
This series is simple, but in more complicated ones it can be useful to have a table to calculate the equivalences of number to letter and vice versa quickly.
We will not be able to take this table with us to do the test, but you will probably have paper to do calculations and on it we can write the equivalence table.
In the example we have seen before, the base series is a fixed factor, but we can find any type of those we saw in the video of numerical series: Fixed or variable factor arithmetic, fixed or variable geometric factors, powers, etc. .
We will see some examples of various types to make it clearer. Try to solve the series that we propose before seeing the solution.
Try to discover the letter that continues this series:
E F H K Ñ?
The resolution of this series is not as evident as in the previous case, so the easiest way to proceed is to obtain the base numerical series.
Using the table that we mentioned before we obtain this base numerical series:
5 6 8 11 15 ?
If we do not see the series factor clearly, it is best to calculate the increments between each two terms in the series:
5 (+1) 6 (+2) 8 (+3) 11 (+4) 15 ?
If we look at the increase we see that we have a series that increases by one unit between every two terms, so the next increase will be (+5).
Thus, the next element of the base series will be 15 + 5 = 20 and if we look in the equivalence table we will see that the position 20 of the alphabet is occupied by the letter "S", so this will be the answer.
Now let's complicate it a little more. Find the letter that continues this series:
O H D B?
In this case we have a decreasing series. The easiest way to proceed is, again, to obtain the base number series:
16 8 4 2 ?
We get the increments between every two terms:
16 (-8) 8 (-4) 4 (-2) 2 ?
In this case we do not have a fixed factor, so it could be an arithmetic series of variable factor or a geometric series.
Let's see if it is a geometric series obtaining the multiplier factor (or divisor) between every two terms of the base series that is: (÷ 2)
We have an arithmetic series in which each element is calculated by dividing the previous one by 2, so The next element of the base series will be: 2 ÷ 2 = 1 and the letter that occupies that position in the alphabet is "A".
Let's see a last example before moving on to the next section:
J S C M V?
This case is somewhat disconcerting since we have one of the letters of the beginning of the alphabet, the "C", in the middle of the series, and on both sides it has letters that are positioned later in alphabetical order so, at first glance, no It is clear whether it is a growing or decreasing series.
We will proceed in the usual way, so we will calculate the base number series:
10 20 3 13 23 ?
Here, the increments of the base series do not give us a clear factor:
10 (+10) 20 (-17) 3 (+10) 13 (+10) 23 ?
In this case, we must remember that the alphabet has a cyclic sequence when solving the series. That is, the next letter after the "Z" will be the "A" that would occupy the "28" position.
Since we see that the factor (+10) appears several times, we will check if the letter "C" is at (+10) positions of the letter "S" and indeed we see that it is.
From "S" to "Z" and then from "A" to "C", there are a total of 10 positions, so by adding (+10) to number 20 we exceed the length of the alphabet so we must subtract 27 (which is the number of letters of the alphabet) to obtain again the valid position of a letter.
In this case 20 + 10 - 27 = 3, which corresponds to the letter "C". With this we have shown that the factor of the series is (+10) so if we add it to the last element of the base series we will have 23 + 10 = 33 and if we subtract 27 we will get 6, which is the position of the letter "F".
With these examples, you can clearly see how to solve this type of series.
If we rely on the equivalence table, we can convert any alphabetical series into a numerical series and solve this with everything learned in the digital series video.
Multiple interleaved alphabetic series
As in the numerical series, it is possible to find two or more nested series in one. This type of series is easy to detect since the length of the series will be longer.
Once we have concluded that we are facing two interleaved series, we will proceed to resolve only the series that affects the solution. Let's see some examples:
C Z D Z F Z G Z I Z J Z L Z?
Here we see that the "Z" is repeated between every two letters so we will have two interleaved series. A very simple one in which the same letter always appears and this one:
C D F G I J L?
When calculating the base series we obtain the following:
C (+1) D(+2) F(+1) G(+2) I(+1) J (+2) L?
The increments are alternately (+1) and (+2), so the next increase will be (+1) and the letter they ask us is therefore the "M".
In this case, one of the series had all its equal terms, (the letter "Z"), but they will not always make it so easy for us. Let's look at a more complicated last example:
T D S E R G Q J P N O?
The length of the series already makes us suspect that it can be two interleaved series, so let's separate them to try to solve them:
Series 1: T S R Q P O
Series 2: D E G J N?
Since the value they ask of us corresponds to the series 2, we can forget the first series (although it seems that it is a simple decreasing series with factor 1).
We calculate the base series of the second, and its increase, and obtain this:
4 (+1) 5 (+2) 7 (+3) 10 (+4) 14 ?
The jump between each two values of the series is increased by one unit so the next increase will be (+5) and the next value of the base series will be 14 + 5 = 19 corresponding to the letter R".
Although not usually very common, we could meet up to three interleaved series. It will be the length of the series that will give us clues about whether it is a multiple series or not.
The mixed series are formed by mixed numerical and alphabetic series. It would be a specific case of the previous section in which one of the series is not alphabetic.
The procedure to solve them would be the same as we explained before. In this case it will be more evident that we are faced with two interleaved series.
Let's see some example:
S 45 X 28 C 11 H 21 M? Q
Here we find several surprises. The first is that the value they ask us for is not the last position.
This can happen and we should not worry. The procedure to follow was already seen in the numerical series video.
What is worrying is that the numerical series is nowhere to take it, and unfortunately the value they ask us for is precisely that sub-series.
The numerical values increase and decrease without any clear criteria, so after a few minutes of frustration trying to solve the series, we will see if both are interrelated, that is, that the values of one depend on the other.
Given the cyclic nature of the alphabetic series, it is possible that the numerical series is based on the positions of the surrounding letters and also becomes a cyclic series.
To verify this, we will substitute the values of each letter for their position in the alphabet and pray for inspiration to arrive:
20 45 25 28 3 11 8 21 13 ? 18
Here, we see that the values of the numerical series grow and decrease as the values of the alphabetic series do, so it is a matter of time that we conclude that the values of the numerical series are calculated by adding the values of the alphabetical series around it: 45 = 20 + 25, 28 = 25 + 3, 11 = 3 + 8, 21 = 8 + 13 and therefore the searched term will be 13 + 18 = 31.
This gives us an idea of the variety of series statements that can be raised.
The only way to successfully overcome any problem of this type is based on practicing everything possible This type of exercise to be able to quickly recognize each case and not waste much time during the real tests.
Alterations and variations
We have already seen how to solve the basic series, which are usually the majority of those we will meet.
On these series, examiners sometimes add some alterations that also affect the result.
These alterations are usually based on the repetition of elements of a series, distinction between vowels and consonants, the use of upper and lower case letters, series of blocks or a combination of all of them.
Let's see some examples:
M N N P Q Q S T T?
If we already have practice with alphabetic series, we can solve most of them without resorting to calculating the base series.
In this case, we clearly see an ascending alphabetical series in which one in two values is repeated.
It is also observed that when a letter is repeated a position in the alphabet is skipped, so the next value will be "V".
Let's look at another case:
O e U i A?
In this example we clearly observe that upper and lower case letters are alternated and that only vowels are being used.
It is a descending series with a jump of a letter between every two terms of the series.
Since it is a cyclic series, the next letter will be a lowercase "o".
It could also be seen as an ascending cyclic series with a factor of +3 and the solution would be exactly the same.
Let's look at a last example in this section:
1AAZ B2BY CC3X?
In this case we have an alphabetical series in blocks that mix numbers and letters. A true gibberish.
Here we have to try to find the logic of the terms of the sequence by seeing the guidelines that follow.
On the one hand, we see that in each block a single number appears, which increases in each term and that moves to the right coinciding with the position it occupies within the block.
Since all terms have the same length of 4 characters, we can deduce that The searched term will look like this: ??? 4.
We can also observe that in each block we have a letter that repeats, that advances in alphabetical order and that is always to the left of the other letter, so The solution should look like this: DD? 4
And finally, we see that the missing letter advances in descending alphabetical order, so the block sought will be: DDW4.
Literal series are based on individual words or sets of words that follow a logical order. From these words the initial one that is used to build the series is normally taken.
Let's see some examples that will make it clearer. Imagine that this series is proposed:
U D T C C S S O?
Since it is a fairly long series, and it does not seem to follow any pattern as a whole, we might think that it is two interleaved series, but after several minutes of unsuccessful efforts, we will have to consider other alternatives.
In this case it is a literal alphabetical series formed by the initials of a set of widely recognizable words that follow an order.
Guess what those words are? This is the solution:
Now it's much clearer, isn't it? The next element of this set of words would be "Nine" and therefore the next letter of the series would be "N".
We propose other typical examples, along with their solution, but you should keep in mind that any set of words that follow an established order can be a good candidate for this type of series.
L M M J V?
In this case it is the days of the week Monday, Tuesday, Wednesday, Thursday, Friday and the next element will be Saturday, so the solution of the series will be "S".
Let's try another series:
E F M A M J?
Have you solved it? Indeed, these are the months of the year: January, February, March, April, May, June, so the letter sought is the "J" of June.
And one last case of this type:
P S T C Q?
That would correspond to the ordinal numbers: First, Second, Third, Fourth, Fifth and the term we are looking for, will be the "S" of Sixth.
In this type of problem it is also possible that you find a series that represents a set of words in reverse order, that is, the first series of this section would become this:
N O S S C C T D?
We go now with another different example. Try to solve this other series:
? T E B A F L A
In addition to the series based on sets of ordered words, we can find others based on a single word.
They tend to represent themselves as the written word itself upside down, although it is also possible to find its messy letters. In this case, if we reverse the order of the series, we have: A L F A B E T?
So the solution would be the letter "O" to form the word "ALPHABET".
Another set of letters widely used in the alphabetical series is that of the Roman numerals: I, V, X, L, C, D, M.
If you thought we had already seen all the existing types of alphabetic series, you are very wrong.
As we already mentioned in the digital series video, the examiners' imagination can create the most varied series, so you have to be open-minded when trying to solve them.
Depending on the academic level of the participants in the test it is possible that you will find series based on the order of the prime numbers, in powers of numbers, in the Fibonacci series, etc.
So, if a series resists you, it is likely that it is not simply based on the numerical order of the letters in the alphabet and you will have to look for alternative resolution methods.
So, finally, we propose a final series for you to squeeze the neurons. Luck!
A A A C E I M M S T?
The truth is that it is a rather complicated example. After trying it as a multiple series, orderly set of words and crumple several sheets of paper, let's see what information we can extract from the series.
We can observe that the letters appear in alphabetical order, but we are unable to find a sequence, either with prime numbers, or with Fibonacci, or with sets of known words, or with the elements of the periodic table, ... so we can think that it is a set of letters that have a meaning as a whole, that is, it's a word.
Since the word is not written in law or vice versa, we conclude that its letters have been reordered, and in what way? Well, in alphabetical order!
So now "only" we have to find a word that contains all the letters of the series including the letter that we must find out. Unless we have a divine inspiration, after several attempts to join pairs of consonant-vocal letters in every imaginable way, Do we get the word MATEMA? ICAS, so we will realize that the letter sought is the "T".
The good news is that you are unlikely to find such complicated series in the psychotechnical tests, and you know that in any case it is advisable to leave the ones that are more difficult for you in the end.
You also have this video entry available:
Good luck in your oppositions!
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