# Number Series – Reasoning Questions

This post is all about number series questions. All the series follows certain pattern, our job is to identify the pattern and calculate the missing number. All the questions are fully solved for full understanding of the concepts

## Find the missing number in the series

### (Q.1)    25, 121, 361, 1369, ?

Options:

(a) 3529         (b) 1723

(c) 3481         (d) 4111

Solution – Here we easily get, it is a square series.
So we choose such an option which is a perfect square

25\quad \quad 121\quad \quad 361\quad \quad 1369\quad \quad \quad 3481\\\ \\{\quad 5 }^{ 2 }\quad { \quad 11 }^{ 2 }\quad \quad \quad { 19 }^{ 2 }\quad { \quad \quad 37 }^{ 2 }\quad \quad \quad \quad { 59 }^{ 2 }\quad

From the above analysis you can see that 3481 is the only number which is perfect square, hence is the answer

### (Q.2) Find the missing number in the following series    24, 340, 2194, 6856, ?

(a) 23892         (b) 24386

(c) 18926         (d) 22445

Here we see that the elements of this series is neither a perfect square nor a perfect cube so here we are doing some subtraction or addition with square or cubes of the numbers which are near to the elements of this series

So here we find, this is ( n3-3) series where n is an alternate prime number that is (3,7,13,19,29)

24\quad \quad \quad \quad \quad \quad 340\quad \quad \quad \quad \quad 2194\quad \quad \quad \quad \quad 6856\quad \quad \quad \quad \quad 24386\\ { 3 }^{ 2 }-3\quad \quad \quad \quad 7^{ 2 }-3\quad \quad \quad \quad \quad { 13 }^{ 2 }-3\quad \quad \quad \quad { 19 }^{ 2 }-3\quad \quad \quad \quad \quad 29^{ 2 }-3

So the last element of the series should be (293-3=24386)

### 03. Find missing number2, 12, 36, 80, 150, ?

(a) 256         (b) 228

(c) 252         (d) 264

After splitting the elements of the series into squares and cubes

Here we get a series of (n3+n2)

where n is 1,2,3,4,5,6

so our last element of the series should be (63+62=252)

\quad 2\quad \quad \quad \quad \quad \quad \quad 12\quad \quad \quad \quad \quad \quad 36\quad \quad \quad \quad \quad \quad 80\quad \quad \quad \quad \quad \quad 150\quad \quad \quad \quad \quad \quad 252\\ { 1 }^{ 3 }+{ 1 }^{ 2 }\quad \quad \quad \quad 2^{ 3 }+2^{ 2 }\quad \quad \quad 3^{ 3 }+{ 3 }^{ 2 }\quad \quad \quad \quad 4^{ 3 }+4^{ 2 }\quad \quad \quad \quad 5^{ 3 }+{ 5 }^{ 2 }\quad \quad \quad \quad { 6 }^{ 3 }+{ 6 }^{ 2 }

Hence 252 is the right answer

### (Q.4)    Find the number6, ?, 62, 123, 214, 341

(a) 11         (b) 25

(c) 36         (d) 47

Solution– Here we get a [(n+1)3-2] series

where n =1,2,3,4,5,6

so our second element  (n=2) should be {(2+1)3-2}

\quad 6\quad \quad \quad \quad \quad \quad \quad 25\quad \quad \quad \quad \quad \quad 62\quad \quad \quad \quad \quad \quad 123\quad \quad \quad \quad \quad \quad 214\quad \quad \quad \quad \quad \quad 341\\\ \\ { 2 }^{ 3 }-2\quad \quad \quad \quad \quad 3^{ 3 }-2\quad \quad \quad \quad { 4 }^{ 3 }-2\quad \quad \quad \quad \quad { 5 }^{ 3 }-2\quad \quad \quad \quad \quad { 6 }^{ 3 }-2\quad \quad \quad \quad \quad { 7 }^{ 3 }-2

Hence 25 is the right answer

### Q.5   Find the missing number

1, 4, ?, 256, 3125

(a) 6           (b) 9 (c) 11         (d) 27

Here we get a nn series

where n is 1,2,3,4,5

\quad 1\quad \quad \quad \quad \quad \quad \quad 4\quad \quad \quad \quad \quad \quad \quad 27\quad \quad \quad \quad \quad \quad \quad 256\quad \quad \quad \quad \quad \quad 3125\quad \quad \quad \quad \quad \quad \\\ \\ \quad { 1 }^{ 1 }\quad \quad \quad \quad \quad \quad 2^{ 2 }\quad \quad \quad \quad \quad \quad \quad 3^{ 3 }\quad \quad \quad \quad \quad \quad \quad 4^{ 4 }\quad \quad \quad \quad \quad \quad \quad \quad \quad { 5 }^{ 5 }

so our 3rd element (n=3) should be 33 =27

### (Q 6)  0, 7, 26,  ?, 124 , 215

(a) 35              (b) 52

(c) 63               (d) 74

Here we see that the elements of the series are near to the cubes of the natural numbers starting from 1,2,3,4,5,6 as we get 1, 8, 27, 64, 125, 216

Here there is a difference of one from our given series with their cubes

\quad 0\quad \quad \quad \quad \quad \quad \quad 7\quad \quad \quad \quad \quad \quad \quad 26\quad \quad \quad \quad \quad \quad \quad 63\quad \quad \quad \quad \quad \quad \quad 124\quad \quad \quad \quad \quad \quad 215\quad \quad \quad \quad \quad \quad \\\ \\ { 1 }^{ 3 }-1\quad \quad \quad 2^{ 3 }-1\quad \quad \quad \quad \quad 3^{ 3 }-1\quad \quad \quad \quad \quad \quad 4^{ 3 }-1\quad \quad \quad \quad \quad { 5 }^{ 3 }-1\quad \quad \quad \quad \quad \quad \quad { 5 }^{ 3 }-1

### (Q 7.)  0 , 2 , 8 , 14 , 24 , 34 , ? , 62

(a) 40             (b) 42

(c) 48               (d) 52

Solution-  Here we see that the elements of the series are near to the squares of the natural numbers starting from 1,2,3,4,5,6 ,7,8

As we get 1, 4, 9, 16, 25, 36, 49, 64

But here we observing that there is difference of one in first element then difference of two with second element and again difference of one with third element with their squares.

It means there is an alternate difference of one and two.

By selecting such an option which gives our seventh element as 72 -1

### (Q 8) 6, 20, 36, 48, 50, ? , 0

(a) 56             (b) 46

(c) 36               (d) 48

Solution- Here we see that the series first is in increasing order then after it is in decreasing order. So here we don’t find any multiplicative or dividing series

But here if we multiply the squares of natural number (1,2,3,4,5,6,7) with the decreasing order of whole numbers (6,5,4,3,2,1,0) we get the required series

6\quad \quad \quad \quad \quad 20\quad \quad \quad \quad \quad \quad 36\quad \quad \quad \quad \quad 48\quad \quad \quad \quad \quad 50\quad \quad \quad \quad \quad \quad 36\quad \quad \quad \quad \quad \quad 0\quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \\\ \\ { 1 }^{ 2 }*6\quad \quad \quad 2^{ 2 }*5\quad \quad \quad \quad 3^{ 2 }*4\quad \quad \quad 4^{ 2 }*3\quad \quad \quad \quad { 5 }^{ 2 }*2\quad \quad \quad \quad 6^{ 2 }*1\quad \quad \quad \quad 7^{ 2 }*0\quad

By selecting such an option which gives our sixth element as (6²×1)

### (Q 9.) 1, 16, 81, 256, 625, ?

(a) 1296                (b) 1225

(c) 2225                (d) 4163

Solution-  Here we easily get a series of power four of naturals numbers (1,2,3,4,5,6)

1 ,        16 ,      81,       256,       625,      1296

(1)4          (2)4        (3)4             (4)4           (5)4              (6)4

By selecting such an option which gives our sixth element as  (6)4

### (Q10) 3, 12, 27 , 48, 75, 108, ?

(a) 162                    (b)183

(c)192                     (d)147

SolutionHere we get all the elements of series are multiples of 3

So we write the elements as

3                4×3              9×7             16×3         25×3          36×3             ?

After simplifying we get series as shown below

3               12               27               48              75              108              147

(1²×3)         (2²×3)         (3²×3)        (4²×3)        (5²×3)        (6²×3)          (7²×3)

By selecting such an option which gives our seventh element as   (7²×3)