# How to solve Missing Numbers Questions: Tips and Tricks

I got many requests from our facebook fans to give some tips and tricks to solve missing numbers questions that are very often asked in SSC, Railway, Banking exams etc. Missing numbers questions are sometimes easy but sometimes take much time to solve them. These questions are the most expected for upcoming SSC Exams 2016 and will help you for your better preparation. We will also play some mock questions/quiz related to this chapter.

Sequence: A Sequence is a set of things (usually numbers) that are in order.

Each number in the sequence is called a term (or sometimes “element” or “member”): ## Finding Missing Numbers

To find a missing number, first I will have to find a Rule behind the Sequence.

Sometimes I can just look at the numbers and see a pattern:

### Example: 1, 4, 9, 16, ?

AnsIr: they are Squares (12=1, 22=4, 32=9, 42=16, …)

Note: I told you they are squares, but if it didn’t hit your mind…you need practice. Now see the following rule.

Rule: xn = n2

Sequence: 1, 4, 9, 16, 25, 36, 49, …

Did you notice how I wrote that rule using “x” and “n” ?

xn means “term number n”, so term 3 is written x3

And I also used “n” in the formula, so the formula for term 3 is 32 = 9. This could be written

x3 = 32 = 9

is not it?

Once I master this Rule I can use it to find any term. For example, the 25th term can be found by “plugging in” 25 wherever n is.

x25 = 252 = 625

See another example:

### Example: 3, 5, 8, 13, 21, ?

Can you solve it? Can you find the missing numbers after 21? If not, then read below.

Could you find any similarity or any sequence?

After the numbers 3 and 5 all the rest are the sum of the two numbers before.

3 + 5 = 8,

5 + 8 = 13 and so on…it is also called as Fibonacci Sequence.

Rule: xn = xn-1 + xn-2

Sequence: 3, 5, 8, 13, 21, 34, 55, 89, …

What is xn-1 here? Actually it just means “the previous term” because the term number (n) is 1 less (n-1).

So, if n was 6, then xn = x6 (the 6th term) and xn-1 = x6-1 = x5 (the 5th term)

So now let’s apply that Rule to the 6th term:

x6 = x6-1 + x6-2

x6 = x5 + x4

I already know the 4th term is 13, and the 5th is 21, so the answer will be….

x6 = 21 + 13 = 34

Pretty simple … just put numbers instead of “n”

## There are many rules….

Finding “the next number” in a sequence are often asked in SSC exams. So let’s take another example.

### What is the missing number in the sequence 1, 2, 4, 7,…. ?

Here are three solutions (there can be more!):

Solution 1: Add 1, then add 2, 3, 4, …

So, 1+1=2, 2+2=4, 4+3=7, 7+4=11, etc…

Rule: xn = n(n-1)/2 + 1

Sequence: 1, 2, 4, 7, 11, 16, 22, …

(That rule looks a bit complicated, but it works)

Solution 2: After 1 and 2, add the two previous numbers, plus 1:

Rule: xn = xn-1 + xn-2 + 1

Sequence: 1, 2, 4, 7, 12, 20, 33, …

Solution 3: After 1, 2 and 4, add the three previous numbers

Rule: xn = xn-1 + xn-2 + xn-3

Sequence: 1, 2, 4, 7, 13, 24, 44, …

So, I have three perfectly reasonable solutions, and they create totally different sequences.

Which is right? They are all right.

And there are other solutions … … it may be a list of the winners’ numbers … so the next number could be … anything!

## Simplest Rule

When in doubt choose the simplest rule that makes sense, but also mention that there are other solutions.

## Finding Differences

Sometimes it helps to find the differences between each pair of numbers … this can often reveal an underlying pattern.

Here is a simple case where we have to find missing numbers after digit 15: The differences are always 2, so I can guess that “2n” is part of the answer.

Let us try 2n:

n: 1 2 3 4 5
Terms (xn): 7 9 11 13 15
2n: 2 4 6 8 10
Wrong by: 5 5 5 5 5

The last row shows that I are always wrong by 5, so just add 5 and I are done:

Rule: xn = 2n + 5

OK, I could have worked out “2n+5” by just playing around with the numbers a bit, but I want a systematic way to do it, for when the sequences get more complicated.

## Second Differences

In the sequence {1, 2, 4, 7, 11, 16, 22, …} I need to find the differences …

… and then find the differences of those (called second differences), like this: The second differences in this case are 1.

With second differences I multiply by “n2 / 2″.

In our case the difference is 1, so let us try n2 / 2:

n: 1 2 3 4 5
Terms (xn): 1 2 4 7 11
n2: 1 4 9 16 25
n2 / 2: 0.5 2 4.5 8 12.5
Wrong by: 0.5 0 -0.5 -1 -1.5

You are close, but seem to be drifting by 0.5, so let us try: n2 / 2 – n/2

 n2 / 2 – n/2: Wrong by: 0 1 3 6 10 1 1 1 1 1

Wrong by 1 now, so let us add 1:

 n2 / 2 – n/2 + 1: Wrong by: 1 2 4 7 11 0 0 0 0 0

The formula n2 / 2 – n/2 + 1 can be simplified to n(n-1)/2 + 1

So by “trial-and-error” I discovered a rule that works:

Rule: xn = n(n-1)/2 + 1

Sequence: 1, 2, 4, 7, 11, 16, 22, 29, 37, …

You can practice many missing numbers questions from

### Verbal & Non Verbal Reasoning by RS Aggarwal

[WpProQuiz 2]

Best wishes!!!

#### 2 Comments

1. Ritu February 11, 2018
2. shiva March 20, 2018