In reasoning section, questions on series are commonly asked. Within series, questions based on alphabets, numerical and alpha-numeric series are generally asked. Talking about alpha-numeric series in particular, also known as alphameric series, is a combination of alphabetic and numeric characters, which seems random but are related to each other by some logic.
Types of Series
Series are classified into three types:
- Numeric series
- Alphabetic series
- Alphanumeric series
Tips to solve Series Completion questions
In order to solve series-based questions, it is important that you:
- Learn the Squares of all the natural numbers from 1 to 25.
- Learn the Cubes of all the natural numbers from 1 to 20.
- Check for the sequence by analyzing the series by checking the difference, by dividing, by checking multiples etc. between the consecutive terms.
- Remember the EJOTY rule. It helps in memorizing the number of alphabet as E corresponds to 5th position, J corresponds to 10th position, O to 15th, T to 20th and Y to 25th position in the alphabets.
Series Completion: Solved Examples
Example 1: A series of letters and numbers is given, the terms of which follow certain definite pattern in group. Find a letter, which should come in the place of blank in this series.
1 A 4 D 7 _ 10 J 13 M
Sol: Above series consists of alternate numerals and alphabets. The number increases by 3 at every step and similarly, alphabet also skips the next two alphabets e.g. 1, 4, 7, 10, 13.Alphabets corresponding to these numerals follow these numbers. So, the letter G will follow 7.
Example 2: Find the next term in the alpha numeric series:
Z1A X2D V6G T21J R88M P445P ?
1. A.N2676S
2. N2676S
3. T2670N
4. T2676N
Sol: The pattern followed is; For the first letter, one alphabet is skipped in the reverse direction i.e. ZXVTRPN. Similarly, third letter skips the next two alphabets i.e. ADGJMPS. Series followed for the numerical values is ×1+1, ×2+2, ×3+3, ×4+4, ×5+5, ×6+6……, so desired numeral is 445 × 6+6 = 2676.The required answer is N2676S.
Example 3: Find the missing term.
2Z5 7Y7 14X9 23W11 34V13 ?
Sol: First number of the series is 2, 7, 14, 23, 34 i.e. +5, +7, +9, +11 now it will be +13 i.e. 34 + 13 = 47. Second term of the series ZYXWV and now it will be U.
Third term of the series 5, 7, 9, 11, 13 and now it will be 15
Thus, 47U15is the required answer.
Example 4: Find the missing terms:
2 3 B _ 6 _ F G _ 5 D _ 8 _ H I
1. C, 7, 4, E, 9
2. D, 8, 6, C, 7
3. E, 8, 7, D, 9
4. W, 8, 7, I, 9
Sol: The given series consists of a group of four characters in which, first two are numerals followed by two alphabets. Alphabet is decided by the number written one place ahead of it in the series. Thus, 2 3 B _ number two corresponds to alphabet B, similarly, the numeral 3 will represent alphabet C, 6 corresponds to F and in missing spot will come the number for alphabet G i.e. 7.
In similar fashion, the whole series can be completed and the right answer will be option A
Example 5: A random alpha-numeric sequence is given. In this series, which of the following is third to the left of fifth from right end?
R * @ 2 D F % ^ E 3 G # 1 Y 9 & R
Sol: First of all, count that the element third to the left of the fifth from the right will appear at position no. 5 + 3 = 8 from the right hand side. Now do the counting and find that the element 8th from the right is the digit 3, hence first option is the answer.
