How to Solve Civil Service Exam Number Series Problems 1

By | December 30, 2013

First of all let me clarify that what you are solving in the Civil Service Examination are number sequences (or letter sequences) and not a number series. A series has a different meaning in mathematics.

Before proceeding with the discussion below, first, try to find the next term in the following sequences.

1. 4, 7, 10, 13, ___

2.  17, 11, 5, -1,  ___

3. C, F, I, L, ___

4. \frac{1}{3}, \frac{5}{6}, \frac{4}{3}, \frac{11}{6}.

Solution and Explanation

Numbers 1 and 2 are the easiest type of sequence to solve. This is because they are integers and you just add (or subtract) a constant number to each term to get the next term.  In solving this type of sequence, you can see this pattern by subtracting adjacent terms (13 – 10 = 3, 10 – 7 = 3, 7- 4 = 3) to see if the difference is constant. If it is, then you will know that you will just have to add the same number to get the next term. Therefore, the next term to the first sequence is 13 + 3 = 16.

 

number sequence

Of course, sequences can also be decreasing. In the second example, the  difference is 6 or it means that 6 is subtracted from a number to get the next term (see Subtraction of Integers). Therefore, the next term is -1 – 6 = -7. 

Screen Shot 2013-12-30 at 10.51.18 AM

 

The third example is composed of letters but the principle is the same: constant difference or constant skips.  C and F, for instance has two letters in between. This is also true between F and I and I and L. Therefore, the next letter in the sequence is O (L, M, N, O).

sequence and series

Fractions and decimals are also included in the sequence problems, so it is important that you master them. In the following example, one half is added each time. As you can see, it is not easy to find the next term of this sequence without manually solving it. The first strategy in solving fraction sequences problems is to subtract the adjacent terms such as

\frac{11}{6} - \frac{4}{3} = \frac{3}{6} which is equal to \frac{1}{2} (since \frac{4}{3} = \frac{8}{6}I

\frac{4}{3} - \frac{5}{6} = \frac{3}{6} which is equal to \frac{1}{2}

\frac{5}{6} - \frac{1}{3} = \frac{3}{6} which is equal to \frac{1}{2} since \frac{1}{3} = \frac{2}{6}.

There is however a better strategy than subtracting the adjacent terms when it comes to sequences on fractions. Sometimes, it is easier to see the pattern if you convert them to similar fractions (fractions with the same denominator). Converting the sequence above to similar fractions gives us

\frac{2}{6}, \frac{5}{6}, \frac{8}{6}, \frac{11}{6}.

From here, it is clear that the next term in the sequence is \frac{14}{6}. Note however that this strategy is best only for sequences with constant difference and may be difficult to use in other types of sequences.

In the next post, we are going to discuss about another type of sequence.

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