# Letter Repeating Series

In the questions on Letter Repeating Series, a series or a bead of alphabets are present. Some of these strings or sequences of alphabets will have missing spaces at regular or irregular places. You will have to complete this sequence in a way that not only makes a regular series but also fits with one of the options. We will also see a shortcut method that will definitely save you a lot of time and guarantee a high rate of success in solving the questions.

## Letter Repeating Series

Let us first start with a series and note down all the parameters that we can get from it. The following is an example of the letter repeating series that you will surely encounter in the IBPS PO, SO, SBI PO, RBI, NABARD exams and other such banking exams.

Example 1: In the following question, a number of letters are given. There are blanks which can be filled with the help of the letters of the options below. Pick the correct option and complete the series:

Q 1: _ aa _ ba _ bb _ ab _ aab

A) aaabb             B) babab              C) bbaab             D) bbbaa

Answer: There are two methods of doing these series which we will discuss in the following sections. Here let us just say that this is how the series looks. The answer to this is the option C). Let us see how we can get there.

### The Easy Way

The easiest way to solve these questions is to put the options back into the blanks. It is a reverse method of solving these questions but it always works. It is fast and will save you a lot of time but it doesn’t always work. We will see in the following section, examples that are of a more difficult nature and won’t be good to solve them this way. But let us see if this method works for the above example. The sequence is  _ aa _ ba _ bb _ ab _ aab. Let us try option A).

For the option A), the series can be written as: aaaa | baab | bbab | baab – which is not any sequence.

For the option B), the series becomes: baaa | babb | baab | baab – which also isn’t a series.

Similarly, you can see for option C) and D). Out of the two, option C) makes a sequence as baab | baab | baab | baab.

Now let us see that other method that we may employ to solve such questions with a 100% accuracy.

Learn the Alphabet Series here in detail.

## The Method For Letter Series

Let us see this method with the help of a few examples.

Example 2: The following letter series has some letters missing from it. Complete the series by using one of the options:

a_bbc_aab_cca_bbcc

A) bacb              B) acba            C) abba            D) caba

Answer: The first step is to count the number of letters that are in the sequence. In the above sequence, this number is 18. Next, we find factors of this number. This gives us the number of ways in which the given series can be represented in a symmetrical way. For example, 18 has the following factors: 2, 3, 6, 9. So we can group the letters either 2 at a time or 3 at a time and so on.

Next step is to find how many different kinds of letters are there. The above series has three types of letters – viz a, b and c. So from the first step, the first number that we check is 3, although it is not necessary that 3 will be the grouping used but highly probable. The series can be written as:

a_b | bc_ | aab | _cc | a_b | bcc

Now we can check the series by inserting the options one by one. The answer here is B) acba and the series becomes:

aab | bcc | aab | bcc | aab | bcc.

Now you might ask if we have to use the options, why not just use the options then? Well, let us see one more example.

### Two Correct Options

Example 3: Complete the series ab_dda__cda__cda:

A) dcdcc             B) ccddc             C) cbcbb           D) cccbb

Answer: You will see that the answer can either be c or d. However, if you start substituting the options, it might get a bit messy. Let us see the method that we have learnt. The series has 16 terms which can be divided into 2, 4, 8 terms taken at once. There are a total of 4 different types of letters. So let us start with groups of 4 letters. The series is:

ab_d | da__ | cda_ | _cda – The series has a double gap which makes it difficult to solve. We see that the option D) doesn’t make anything click. The correct option hence is C) cbcbb and the series is abcd | dabc | cdab | bcda which is a sequence.

Learn the Alpha-Numeric Series here.

## Practice Questions

Q 1: Complete the series by using the options given below:

cc_ccdd_d_cc_ccdd_dd

A) dcdcc            B) dcddc            C) dccdd              D) ddddd

Ans: B) dcddc

Q 2: _bca_cs_c_b_

A) aabbc             B) abbbc           C) aabcc             D) abbac

Ans: D) abbac

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### One response to “Alphabet Series”

1. Alan Livesey says:

In example 2:
“You can see that each of the numbers is a square and that the sequence is a perfect square series. 1, 22, 32, 42 (=16). The alphabet that corresponds to 16 is P. ”
“You can see that each of the numbers is a square and that the sequence is a perfect square series. 1, 2×2, 3×3, 4×4 (=16). The alphabet that corresponds to 16 is P. ”
(I suspect that the original text from which this was prepared used superscript but this has not been reflected in the online version, so use “2×2” or “2^2” which most people are familiar with from Excel)

In Circular Arrangement Series:
“These type of questions are similar to the ones we saw earlier. But there our numbering scheme would stop at 26 with X.”
“These type of questions are similar to the ones we saw earlier. But there our numbering scheme would stop at 26 with Z.”

In example 3:
“Answer: The lesser the number of alphabets present, the greater the difficulty of the question. Here you see that V and A have a difference of 4 alphabets between them. Similarly, A and H have a difference of 6 alphabets between them if we follow the circular order of the alphabets. Thus the next alphabet will have to have a difference of 8 alphabets with H. This alphabet is Q. Thus the series is V, A, H, Q. Therefore the correct option is s) P.”
“Answer: The lesser the number of alphabets present, the greater the difficulty of the question. Here you see that V and A have a difference of 4 letters between them. Similarly, A and H have a difference of 6 letters between them if we follow the circular order of the alphabets. Thus the next alphabet will have to have a difference of 8 letters with H. This letter is Q. Thus the series is V, A, H, Q. Therefore the correct option is d) P.”

In example 4:
“Answer: We will have to figure out the rule to every sequence. If you use the table, you will see that it becomes much more convenient to guess the rule. For example, in the first series, Q = 17; T = 20, X = 24; C = 29 [circular alphabet order]. Thus it forms a series under the rule. Similarly for the second option, F 6, P = 16, Z = 26 and J = 36. It also forms a correct sequence. Let us see the third one i.e. W = 23; U = 21; R = 18; and N = 14. So it is a wrong sequence. In place of N = 14, we should have had O.

That means the only series here that has a wrong term should be d). Let us check it. We have A = 1, L = 12, W = 23, H = 34.
Answer: We will have to figure out the rule to every sequence. If you use the table, you will see that it becomes much more convenient to guess the rule. For example, in the first series, Q = 17; T = 20, X = 24; C = 29 [circular alphabet order]. Thus it forms a series under the rule. Similarly for the second option, F 6, P = 16, Z = 26 and J = 36. It also forms a correct sequence. Let us see the fourth one i.e. A = 1, L = 12, W = 23, H = 34 which is a correct sequence.

Let us see the third one i.e. W = 23; U = 21; R = 18; and N = 14. So it is a wrong sequence. In place of N = 14, we should have had O.

Kind regards,
Alan

PS If you have other material that needs proof-reading I am frequently called upon to spot typos and grammatical errors in texts of all types including dense technical ones.