Tips & Tricks to Solve Alphabet Series Problems
Here in this article, we will discuss alphabetical series tricks.
Two or three questions from alphabetic series is there in every exam. In such questions of alphabetic series, if the question consists of a single alphabet series, then you have to solve the logic implied in the sequence. Then fill in the missing character with a correct choice. Here in this article, we will discuss alphabetical series tricks.
Remember all the alphabets and their place number ,You can even note down during your exam.
There are 26 letters in English alphabets.
The first half of alphabet series is from A to M and the half is from N to Z.
A-M – 1-13 (First Alphabetical Half)
N-Z –14-26 (Second Alphabetical Half)
So if you remember these intervals,It would be helpful in solving alphabet series problems.
Various types of Alphabetical series
Type – I: ALPHABET SERIES
Increasing by a definite number
For Example: IJKL? ( each letter increases by 1)
AGMSY? ( each letter increases by 6 place to its right position)
Decreasing by a definite number
For Example: ZXVTRP ? ( each letter decreases by 2 places to its left )
For Example: DEGJNS? ( +1,+2,+3,+4,+5)
For Example: ZYWTP ( -1,-2,-3,-4 ..)
ZTOKHFE ( -6,-5,-4,-3,-2,-1)
Decreasing and Increasing by a constant value.
For Example: DFCEBDACZ (+2,-3,+2,-3,…)
TYPE-II : ALPHANUMERIC SERIES
EX-1: Z1A, X2D,V6G,T21J,R88M, P445P,?
First letter: ZXVTRP (-2,-2,-2,…..)
Second letter: ADGJMP ( +3, +3,+3,…)
Series of numerals: 1,2,6,21,88,445 ( x1+1, x2+2, x3+3…)
So next term is N2676S.
First numeral- 2,7,14,23,34 (+5,+7,+9,+11..)
Second letter- ZYXWV ( decreases by 1 each time)
Third numeral- 5,7,9,11,13 ( increases by 2 each time)
EX-3 : W-144 , U-121, S-100, Q-81,?
First letter- decreases by 2 each time
Second numeral- square of 12,11,10,9,8..
TYPE-III : CONTINUOUS PATTERNS SERIES
Ex-1 : ab_ _ baa_ _ ab_
Options: i) aaaaa ii) aabaa iii) caabab iv) baabb
Solution: our answer is ii) . Here series aba is repeated
Options: i) abba ii) baab iii) aabb iv) abab
Solution: our answer is ii) . The series is abb/aaabbb/aaaabbbb/a. Thus the letter are repeated twice , then thrice , then four times and so on .
Ex.3 – _bc_ca_aba_c_ca
Options i)abcbb ii)bbbcc iii)bacba iv)abbcc
Solution: our answer is i) . The series is abc/bca/cab/abc/bca. Thus the letter change in cyclic order .
Options: i) adabcd ii) bdbcba iii) cdbbca iv)daabbc
Solutions: our answer is i). The series is acdb/dacb/cdab/acdb/da. Each group of four letters contains the letters of the previous group in the order – third , first , second and fourth.
Options: i) aabba ii) bbaab iii)abaaa iv)baabb
Solutions: our answer is iii). The series is aabbbb/aaabbb/aaaa. At each step , the number of a’s increases by one while the number of b’s decrease by one.