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## Ten most expected missing number series questions for SBI PO with solution for other competitive Exams

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# Missing  number series questions for SBI PO with solution

Ten most expected  missing  number series questions for SBI PO with solution , for SSC GL , SSC CHSL , RRB NTPC and other competitive Exams  have been discussed in this post.  These types of series questions are asked in many other competitive exams like SI, CPO and various entrance exams.

## Problem # 1

Take the difference of two consecutive terms of the given series. This difference will be the pair of same odd numbers in every case.
16 - 5 = 11 (The difference of 1st and 2nd number is pair of odd number one)
49 - 16 = 33  (The difference of 2nd and 3rd number is pair of odd number three )
104 -  49 = 55  (The difference of 3rd and 4th number is pair of odd number  five)
? - 104 = 77   ( Similarly the difference of 4th and 5th number will pair of odd number seven)
⇒ ? =  77 + 104  (And in the same way the difference of 5th and 6th number is pair of odd number nine )
⇒ ? = 181
280 - ? = 99
⇒ ? = 280 - 99
⇒? = 181

Option (2)181 will be correct option

## In this series every number is written as power of one less than the term,s position and taking as base of  5 . i.e. 1st term is written as power of 0 , 2nd term is written as power of 1, 3rd term is written as power of 2 , 4th term is written as power of 3 and so on.
5⁰ - 0 = 1 - 0 = 1  (Power 0 of base 5 - same power )
5¹ - 1 = 5 - 1 = 4   (Power 1 of base 5  - same power   )
5² - 2 = 25 - 2 = 23   (Power 2 of base 5  - same power  )
5³ - 3 = 125 - 3 = 123   (Power 3 of base 5  - same power )
5⁴ - 4 = 625 - 4 = 621   (Power 4 of base 5  - same power  )
Option (4)621 is correct option

## In this series every number is written as power of one greater than the term,s position and taking as base of  5 . i.e. 1st term is written as power of 0 , 2nd term is written as power of 1, 3rd term is written as power of 2 , 4th term is written as power of 3 and so on.5⁰ + 0 = 1 + 0 = 1  (Power 0 of base 5 + same power )5¹ + 1 = 5 + 1 = 6   (Power 1 of base 5  + same power   )5² + 2 = 25 + 2 = 27   (Power 2 of base 5  + same power  )5³ + 3 = 125 + 3 = 128   (Power 3 of base 5  + same power )5⁴ + 4 = 625 + 4 = 629   (Power 4 of base 5  + same power  )

Option (2)629 will be correct option.

## Problem # 4

In this series every number is written as cube of  the term,s position and one number less than it . i.e. 1st term is written as cube of 1 and less than 1 , 2nd term is written as cube of 2 and less than 1, 3rd term is written as cube of 3 and one lees than it , 4t term is written as cube of 4 and one less than it and so on.

1³ - 1 = 1 - 1 = 0
2³ - 1 = 8 - 1 = 7
3³ - 1 = 27 - 1 = 26
4³ - 1 = 64 - 1 = 62
5³ - 1 = 125 - 1 = 124
6³ - 1 = 216 - 1 =  215
Option (1)124 will be correct option.

## (5  × 1)  + 2 = 5  +  2 = 7 ( To get 2nd term Multiplying 1st term with 1 and add 2 to it)

(7 × 2)  -  4 = 14 - 4  = 10 ( To get 3rd term Multiplying 2nd term with 2 and subtract 4 to it ).
(10 × 3) + 6 = 30 + 6 = 36 ( To get 4th term Multiplying 3rd term with 4 and add 6 to it)
(36 × 4)  - 8 = 144 - 8 = 136 ( To get 5th term Multiplying 2nd term with 2 and subtract 4 to it )
(136 × 5)  + 10 = 680 +10= 690 ( To get 2nd term Multiplying 1st term with 1 and add 2 to it)
Option (2)690 will be correct option.

## Problem # 6

To get 2nd term ,multiply 1st term with half and add cube of 1 to it .
(12 × 0.5 ) + 1 = 6 + 1 =  6 +  1³ = 6 + 1 = 7
To get 3rd term ,multiply 2nd term with one ( double of half ) and add cube of 2 to it .
(7 × 1)  + 8 = 7 +  2³ =  7 + 8 = 15
To get 4th  term ,multiply 3rd term with 2 ( double of 1 ) and add cube of 3  to it .
(15 × 2) +  27 =  30 +  3³ = 57
To get 5th  term ,multiply 4th term with 4 ( double of 2 ) and add cube of 4  to it .
( 57 × 4 ) + 64 =  228 +  4³ = 292
To get 6th  term ,multiply 5th term with 8 ( double of 4 ) and add cube of 3  to it .
( 292 × 8 ) + 125 = 2336 + 125 = 2461
option (4)2461 is correct option

## Problem # 7

Multiply 1st term of the series with one and half  (1.5) to get 2nd term of the series.
10 × 1.5 = 15
Multiply 2nd term of the series with 3  (  double the 1.5 i.e.  2 × 1.5 = 3 ) to get 3rd term of the series.
15 × 3 = 45
Multiply 3rd term of the series with  6 (  double the 3 i.e. 3 × 2 = 6) to get 4th  term of the series.
45 × 6 = 270
Multiply 4th term of the series with 12  ( double the 6 i.e. 6 × 2 = 12 ) to get 5th term of the series.
270 ×  12  = 3240
Multiply 5th term of the series with 24 (  double the 12 i.e. 12 × 2 = 24 ) to get 2nd term of the series.
3240 × 24 =  77760
Option (2)77760 is correct option

## (2 × 5) - 1 = 10 - 1 = 9  ( 2nd term is written as multiple of 5 with 1st term  and decrease it by 1 )
(9 × 5) + 3 =  45 + 3 = 48  ( 3rd term is written as multiple of 5 with 2nd term  and increase it by 3)
(48  × 5) - 5 = 240 - 5 =235  ( 4th term is written as multiple of 5 with 3rd term  and decrease it by 5)
( 235 × 5 ) + 7 = 1185 + 7 = 1182  ( 5th term is written as multiple of 5 with 4th term  and decrease it by 7 )

Note carefully in this problem ,operation of decreasing and increasing used alternatively.
Option (2)1182 is correct option
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## Problem # 9

In this series 1st term is 12 , To get 2nd term from it we just multiplied it with half
12 × 1/2 =  6

To get 3rd term from 2nd term we just multiplied it with double of the number what we have multiplied in last term i.e. 2 × ( 1/2) =1
6  × 1 = 6
To get 4th term from 3rd term we just multiplied it with double of the number what we have multiplied in last term i.e. 2 × ( 1) =2
6 × 2 = 12
To get 5th term from 3rd term we just multiplied it with double of the number what we have multiplied in last term i.e. 2 × ( 2 ) =4
12 × 4 = 48
To get 6th term from 5th term we just multiplied it with double of the number what we have multiplied in last term i.e. 2 × ( 4 ) =8
48 × 8 = 384
Hence (2)384 will be correct option

To find out the solution of  above series problem just add all the digits of the every number given , this sum will leads to the next number of the series
5 + 3 + 4 + 2 = 14 ( 2nd number in series)
Similarly add all the digits of the 3rd number, this sum will leads to the 4th number of the series
5 + 2 + 3 + 1 = 11 ( 4th number in series)
4 + 1 + 2 + 0 = 17 ( 6th number in series)
6 + 7 + 3 + 2 = 18  ( last/ Missing number in series)
Option (4)18 is correct option

Ten most important Missing number series questions and answers, Number Series Questions for SBI Clerk ,SBI PO with solution  for SSC GL , SSC CHSL , RRB NTPC and other competitive Exams were discussed in this post.  comment your valuable suggestions regarding this post and for further improvement.

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