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Friday, July 31, 2020
Series of Alphabets: Rules
Series of Alphabets: Rules
Questions related to series of Alphabets can be asked based on certain rules:
Rule 1: Series can have forward counting numbers from A = 1, z = 26
A B C D E F G H I J K L M N O P Q R S T U V W X Y Z
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26
(Left) ------------------------------------------------------------------------------------------ (Right)
This is English alphabetical order.
Rule 2: Series can have Reverse counting numbers from A=26, Z = 1
Z Y X W V U T S R Q P O N M L K J I H G F E D C B A
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26
(Left) ------------------------------------------------------------------------------------------- (Right)
This is Reverse or Opposite English alphabetical order.
Rule 3: There are two haves in English alphabets:
A B C D E F G H I J K L M N O P Q R S T U V W X Y Z
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26
(Left) ------------------------------------------------------------------------------------------ (Right)
(a) First Half in alphabetical order from left
A B C D E F G H I J K L M
1 2 3 4 5 6 7 8 9 10 11 12 13
(b) Second half in alphabetical order from left
N O P Q R S T U V W X Y Z
14 15 16 17 18 19 20 21 22 23 24 25 26
(c) First Half in alphabetical order from right
Z Y X W V U T S R Q P O N
26 25 24 23 22 21 20 19 18 17 16 15 14
(d) Second Half in alphabetical order from right
M L K J I H G F E D C B A
13 12 11 10 9 8 7 6 5 4 3 2 1
Rule 4: There are two haves in Reverse English alphabets:
Z Y X W V U T S R Q P O N M L K J I H G F E D C B A
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26
(Left) ------------------------------------------------------------------------------------------- (Right)
(a) First Half in reverse alphabetical order from left
Z Y X W V U T S R Q P O N
1 2 3 4 5 6 7 8 9 10 11 12 13
(b) Second half in reverse alphabetical order from left
M L K J I H G F E D C B A
14 15 16 17 18 19 20 21 22 23 24 25 26
(c) First Half in reverse alphabetical order from right
A B C D E F G H I J K L M
26 25 24 23 22 21 20 19 18 17 16 15 14
(d) Second Half in reverse alphabetical order from right
N O P Q R S T U V W X Y Z
13 12 11 10 9 8 7 6 5 4 3 2 1
Alphabet Types:
Vowel – a e i o u (5 letters)
1 5 9 15 21 = 51
+4 +4 +6 +6
Consonant – b c d f g h j k l m n p q r s t v w x y z (21 Letters)
Opposite Letters in English Alphabet
1 2 3 4 5 6 7 8 9 10 11 12 13
A B C D E F G H I J K L M
↕ ↕ ↕ ↕ ↕ ↕ ↕ ↕ ↕ ↕ ↕ ↕ ↕
Z Y X W V U T S R Q P O N
26 25 24 23 22 21 20 19 18 17 16 15 14
Series Examples
Series: B E I N T ? (Series of one letter)
(a) W (b) Y (c) A (d) C
Solution:
+3 +4 +5 +6
B → E → I → N → T
So,
Next number will be +7 after T i.e.
U V W X Y Z A
1 2 3 4 5 6 7
(Remember after Z,series continues again from A)
So, Answer is A
Series: AZ CX
FU JQ ? (Series of two letter)
(a) OL (b) MK (c) OK (d) ML
Solution:
First Letter
+2 +3 +4
A → C → F → J
So, Next letter will be J + 5 = K L M N O , So, next first letter is O.
Second Letter
- 2 - 3 - 4
Z → X → U → Q
So, Next letter will be Q - 5 = P O N M L , So, next first letter is L.
So the next term of the series will be OL. Answer is OL
Series of Numbers: Rules
Series of Numbers
Perfect Square Numbers:
12 = 1 22 = 4 32 = 9 42 = 16 52 = 25
62 = 36 72 = 49 82 = 64 92 = 81 102 = 100
112 = 121 122 = 144 132 = 169 142 = 196 152 = 225
162 = 256 172 = 289 182 = 324 192 = 361 202 = 400
212 = 441 222 = 484 232 = 529 242 = 576 252 = 625
262 = 676 272 = 729 282 = 784 292 = 841 302 = 900
Examples
Series: 4 9 16 25 ? 49 64
Solution
In this series, we can see the pattern is perfect squares of 2, 3, 4, 5, .............
So next perfect square will be 6 × 6 = 36
Series: 63 80 99 120 143 ? 195
Solution:
See 63 = 64 - 1 = 8 × 8 - 1
80 = 81 - 4 = 9 × 9 - 1
99 = 100 - 4 = 10 × 10 - 1
120 = 121 - 4 = 11 × 11 - 1
143 = 144 - 4 = 12 × 12 - 1
See 63 = 64 - 1 = 8 × 8 - 1
80 = 81 - 4 = 9 × 9 - 1
99 = 100 - 4 = 10 × 10 - 1
120 = 121 - 4 = 11 × 11 - 1
143 = 144 - 4 = 12 × 12 - 1
We can see that the pattern is a × a - 1
So next number will be 13 × 13 - 1 = 169 - 1 = 168
Series: 125 104 85 68 ? 40
Solution
See 125 = 121 + 4 = 11 × 11 + 4
104 = 100 + 4 = 10 × 10 + 4
85 = 81 + 4 = 9 × 9 + 4
68 = 64 + 4 = 8 × 8 + 4
We can see that the pattern is a × a + 4
So next number will be 7 × 7 + 4 = 49 + 4 = 53
Perfect Cube Numbers:
13 = 1 23 = 8 33 = 27 43 = 64 53 = 125
63 = 216 73 = 343 83 = 512 93 = 729 103 = 1000
113 = 1331 123 = 1728 133 = 2197 143 = 2744 153 = 3375
Examples
Series: 8 27 64 125 ? 343 512
Solution:
In this series, we can see the pattern is perfect cubes of 2, 3, 4, 5, .............
So next perfect cube will be 6 × 6 × 6 = 216
Series: 63 124 215 342 ? 728
Solution:
See 63 = 64 - 1 = 4 × 4 × 4 - 1
124 = 125 - 1 = 5 × 5 × 5 - 1
215 = 216 - 1 = 6 × 6 × 6 - 1
342 = 343 - 1 = 7 × 7 × 7 - 1
We can see that the pattern is a × a × a - 1
So next number will be 8 × 8 × 8 - 1 = 512 - 1 = 511
Series: 1004 733 516 347 ? 129
Solution:
See 1004 = 1000 + 4 = 10 × 10 × 10 + 4
733 = 729 + 4 = 9 × 9 × 9 + 4
516 = 512 + 4 = 8 × 8 × 8 + 4
347 = 343 + 4 = 7 × 7 × 7 + 4
We can see that the pattern is a × a × a + 4
So next number will be 6 × 6 × 6 + 4 = 216 + 4 = 220
Geometric Progression
In these series, first we check for the multiplicand r by dividing preceding number from the previous number in the series. We know that GP series is formed in the following manner.
a ar ar2 ar3 ar4 ……………………..arn-2 arn-1
Example:
Series: 1 2 4 8 16 32 64 ? 256 512
Solution:
Divide 2 / 1 = 2, 4 / 2 = 2, 8 / 4 = 2, .... and so on..... So we can say that,
This is a series with a = 1 and r = 2
So series will be
1 × 2 = 2, 2 × 2 = 4, 4 × 2 = 8, 8 × 2 = 16, 16 × 2 = 32,
32 × 2 = 64, 64 × 2 = 128
So the missing number will be 128
The same can be found using formula: ar6 - 1 = ar7 = 1 × (2)7 = 128
Series: 64 32 16 8 4 2 1 ?
Solution:
Divide 32 / 64 = 1/ 2, 16 / 32 = 1/ 2, 8 / 16 = 1 / 2, .... and so on..... So we can say that,
This is a series with a = 64 and r = 1/2
So series will be
64 × 1/2 = 32, 32 × 1/2 = 16, 16 × 1/2 = 8, 8 × 1/2 = 4, 4 × 1/2 = 2,
2 × 1/2 = 1, 1 × 1/2 = 1/2 = 0.5
So the missing number will be 1/2 or 0.5
The same can be found using formula: ar6 - 1 = ar7 = 64 × (1/2)7 = 64 / 128 = 1 / 2 = 0.5
Series: 3 4 6 10 18 ? 66 130 258
Solution:
This is a series with a = 3 and r = 2 with a difference of 2 as follows
3 × 2 - 2 = 6 - 2 = 4, 4 × 2 - 2 = 8 - 2 = 6, 6 × 2 - 2 = 12 - 2 = 10,
10 × 2 - 2 = 20 - 2 = 18 18 × 2 - 2 = 36 - 2 = 34 34 × 2 - 2 = 68 - 2 = 66
So the missing number will be 34
Arithmetic Progression:
In these series, first we check for the difference d by subtracting previous number from the preceding number in the series. We know that AP series is formed in the following manner.
a a + d a + 2d a + 3d …………….. a + (n-2)d a + (n-1)d
Example:
Series: 4 9 14 19 ? 29 34 39
Solution:
Subtracting 9 - 4 = 5, 14 - 9 = 5, 19 - 14 = 5, .... and so on..... So we can say that,
This is a series with a = 4 and d = 5
4 + 5 = 9, 9 + 5 = 14, 14 + 5 = 19, 19 + 5 = 24, 24 + 5 = 29, ....
So the missing number will be 24
The same can be found using formula: a + (n - 1)d = 4 + (5 - 1)× 5 = 4 + 20 = 24
Series: 37 30 23 16 9 2 ? -12
Solution:
Subtracting 30 - 37 = - 7, 23 - 30 = - 7, 16 - 23 = -7, .... and so on.... So we can say that,
This is a series with a = 37 and d = - 7
37 - 7 = 30, 30 - 7 = 23, 23 - 7 = 16, 16 - 7 = 9, 9 - 7 = 2, 2 - 7 = - 5
So the missing number will be (- 5)
The same can be found using formula: a + (n - 1)d = 37 + (7 - 1)× (- 7) = 37 - 42 = - 5
Arithmetic Progression with variable Difference (d):
In these series, first we check for the difference 'd' by dividing preceding number from the previous number in the series. We know that GP series is formed in the following manner.
We will see that the values of 'd' in each calculation is different but having some definite pattern. we can understood it using some examples.
Example:
Series: 2 5 9 14 20 27 ?
Solution:
Subtracting d = 5 - 2 = 3, d = 9 - 5 = 4, 14 - 9 = 5, 20 - 14 = 6, 27 - 20 = 7
See the sequence of d = 3 3 + 1 = 4 4 + 1 = 5 5 + 1 = 6 6 + 1 = 7
So next d will be 7 + 1 = 8
So next value in the series will be 27 + 8 = 35
Geometric Progression with variable Multiplicand (r):
In these series, first we check for the multiplicand r by dividing preceding number from the previous number in the series. We know that GP series is formed in the following manner.
We will see that the values of 'r' in each calculation is different but having some definite pattern. we can understood it using some examples.
Example:
Series: 2 6 24 120 720 ?
Solution:
Divide: r = 6 / 2 = 3, r = 24 / 6 = 4, r = 120 / 24 = 5, r = 720 / 120 = 6
See the sequence of r = 3 3 + 1 = 4 4 + 1 = 5 5 + 1 = 6
So next r will be 6 + 1 = 7
So next value in the series will be 720 × 7 = 5040
Prime numbers:
These are the numbers which are either divisible by 1 or itself, means it has only and necessarily two factors.
1 is not a prime number because it has only one factor that is 1.
2 is only even prime number which has two factors that is 1 and 2 itself.
2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89, 97, 101, 103, 107, 109, 113, 127, 131, 137, 139, 149, 151, 157, 163, 167, 173, 179, 181, 191, 193, 197, 199
Series: 5 7 11 13 17 ?
Solution:
We can see that this is the series of prime numbers. So next number will be 19
Series: 3 7 13 19 29 ?
Solution:
We can see that this is the series of alternate prime numbers.
3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37
Taking alternate numbers we get 3 7 13 19 29 37
So next number will be 37
Series: 26 34 38 46 58 ?
Solution:
We can see that there is neither AP nor GP or any other theory in this series. but all numbers are even numbers.
Divide all numbers by 2, and we get
26 / 2 = 13, 34 / 2 = 17, 38 / 2 = 19, 46 / 2 = 23, 58 / 2 = 29
Here we can see the hidden prime number series 13, 17, 19, 23, 29
Next prime number will be 31
So next number in the series will be 31 × 2 = 62
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