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Reasoning - Sequence and Series

Updated on November 23, 2017
Every next group of sticks increasing with a common difference of 1 - a simple AP
Every next group of sticks increasing with a common difference of 1 - a simple AP
Each group is increasing in multiple of 2 - a simple GP
Each group is increasing in multiple of 2 - a simple GP

Number Sequence and Series

Sequence: A sequence is an ordered list of numbers or a sequence is a set of numbers arranged in a particular order.

Series: A series is composed of a sequence of terms that are added up.

Let us start with few simple examples of number sequences.

  • Example 1: 1,2,3,4,5,6,7,8,....
  • Example 2: 7,10,13,16,19,22,.....
  • Example 3: 12,17,22,27,32,37,.....

  • Example 4: 1,2,4,8,16,32,64,128,.....
  • Example 5: 3,9,27,81,243,749,2247,.....
  • Example 6: 1/2,1/4,1/8,1/16,1/321,1/64,....

  • Example 7: 1,1/2,1/3,1/4,1/5,1/6,1/7,1/8,....
  • Example 8: 1/7,1/10,1/13,1/16,1/19,...
  • Example 9: 1/12,1/17,1/22,1/27,....

Above mentioned examples are mainly in three mathematical serquences. They are AP, GP, and HP. AP stands for arithmetic progression, GP stands for geometric progression, and HP stands for harmonic progression. First three (1,2,3) examples are in AP, second three examples (4,5,6) are in GP, and the last three examples (7,8,9) are in HP. Let us try to understand the three sequences one by one.

Arithmetic Progression (AP):

Arithmetic Progression is also called arithmetic sequence. In a sequence of AP, the difference between any two consecutive numbers or terms is constant and this constant, we are denoting as d, is called common difference. If first or last term of the series is known and we know the common difference, we can get the next term or any term of the series. Even, we can calculate the sum of, say n number of terms. Many questions related to AP are asked in examination. Let us explain in detail with an example.

An arithmetic progression is given by the formula:

a, (a+d), (a+2d), (a+3d), (a+4d), (a+5d),+.......[a+(n-1)d], ......

where a = first term, d = common difference


4, 6, 8,10,12,14,..... is an arithmetic progression with a = 4 and d = 2

13,18, 23, 28, 33,.... is an arithmetic progression with a = 13 and d = 5

Note: If the three terms a, b, c are in AP then,

b = (a+c)/2


2b = a + c

nth term of an arithmetic progression:

a, (a+d), (a+2d), (a+3d), (a+4d), (a+5d),+.......+ [a+(n-1)d], .....

is given by the formula:


tn = a + (n – 1)d = l

where tn = nth term, a = the first term , d = common difference, l = the last term

Sum of n terms in an arithmetic series:


Sn = n/2 [ 2a + (n-1)d ] = Sn = n/2( a + l )


where, a = first term of the series

n = number of terms in a series

d = common difference

tn = nth term of the series

Sn = Sum of n terms of the series

l = last term of the series

To find the sum of an AP series when the first term (a),the common difference (d), and the number of terms (n) are known.


Find sum of the first 100 terms of the series?

1 + 2 + 3 + 4 + 5 + 6+............................100 + 101 +.....

solution: Here, a = 1, d = 1, n = 100, so

S100 = 100/2[ 2×1 + (100 - 1)×1 ]

S100 = 50[ 2 + 99 ]

S100 = 50[101]

S100 = 5050

To find the sum of an AP series when the first term (a), the common difference (d), and the last term (n) are known.

Example 2:

Q. Find the sum of the series:

5 + 10 +15 +20 +25 +............+60

Solution: Here, a = 5, l = 50, d =10 - 5 = 20 -15 = 5 (take any two consecutive terms, you will get the same difference)

n = ?, Sn = ?

To fins the sum first, we will have to find n (number of terms)

Using formula,

l = a + (n - 1)d

60 = 5 + (n - 1)5

55 = 5n - 5

5n = 55+5

5n = 60

n = 12 (dividing both the sides by 12)

Now, using sum formula

Sn = n/2[a + l]

Sn = 12/2[5 + 60]

Sn = 6[65]

Sn = 390

Arithmetic Mean

If a, b, c are in AP, b is the Arithmetic Mean (AM) between a and c.

In this case,

b = (a+c)/2

2b = a+c

Note: If a, a1, a2 ... an, b are in AP we can say that a1, a2, a3, .... an are the n Arithmetic Means between a and b.

Geometric Progression:

Geometric Progression (GP) or Geometric Sequence is sequence of non-zero numbers in which the ratio of any term and its preceding term is always constant. In other words, it is a sequence of numbers where each term after the first is found by multiplying the previous one by a fixed, non-zero number called the common ratio (r).

A geometric progression (GP) is given by the formula:

a, ar, ar2, ar3, ...arn−1,.....

where a = the first term, r = the common ratio, n = number of terms.


  • 3, 6, 12, 24, 48, ... is a geometric progression (GP) with a = 3 and r = 2
  • 1, 2, 4, 8, 16, ... is a geometric progression (GP) with a = 1 and r = 2

Geometric Mean:

If a, b, c are in GP then,

b2 = ac



b = √ac


where b is the geometric mean between a and c

nth term of a geometric progression (GP):

a, ar, ar2, ar3, ...arn−1,.....


tn = arn−1


where tn = nth term, a = the first term , r = common ratio, n = number of terms

Sum of n terms of a geometric series:

Let a geometric series be represented by the formula:

a + ar + ar2 + ar3 + ar4 + .........arn-1......, where a = first term and r = common ratio, then

Example 1
Find the 10th term in the series 2, 4, 8, 16, 32,....

a = 2, r = 4/2 = 2, n = 10

10th term,

t10 = arn−1=2×210−1= 2×29 = 2×512 =1024 →Ans.

Example 2
Find 6th term in the series 5, 15, 45, 60, ...

a = 5, r = 15/5 = 3, n = 6

6th term,

t6 = arn−1 = 5×36−1 = 5×35= 5×243 = 1215 →Ans.

Sum of Infinite Geometric Series

A finite sum can be obtained from GP with infinite terms if and only if -1.0 ≤ r ≤ 1.0 and r ≠ 0.

Harmonic Progression

The reciprocals of the corresponding terms of an AP, containing non-zero terms, constitute an HP i.e. if 1, 2, 3, 4, 5, ....... are in AP, then 1, 1/2, 1/3, 1/4, 1/5,..... are in HP. But if the sequence is 0, 1, 2, 3 ,4, 5 ..... , the reciprocals of its corresponding terms will not constitute an HP. Reciprocal of zero is not defined.

nth term of an H.P.

If a, a + d, a + 2d, a + 3d,............ a + (n-1)d are in AP,

where nth term of AP = a + (n - 1)d,

then 1/a, 1/(a + 2d),1/(a + 3d),........1/[a + (n-1)d] are in HP.



nth term of HP = 1/[a + (n-1)d]



Q.1 Find the 16th term of the given HP.

1, 1/3, 1/5, 1/7,.......

Solution: Using formula,

tn = 1/[a + (n - 1)d]

t16 = 1/[1 + (16 - 1)2]

t16 = 1/[1 + 30]

t16 = 1/31 →Ans.

Harmonic Mean: If a, b, c are in HP and a ≠ 0, b ≠ 0, c ≠ 0, then

1/a, 1/b, 1/c are in AP

→ 1/b = (1/a + 1/c)/2


→ b = 2ac/(a + c)


where b is the harmonic mean between a and b

Try, what you have learned.

Just go through this quiz and judge yourself.

Number Series test 1

view quiz statistics

© 2016 sonal


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    • profile image


      4 months ago


    • verdict profile image

      George Dimitriadis 

      14 months ago from Templestowe

      You have included many good points in your article.

      For the infinite sum, the value of r, the common ratio, is strictly BETWEEN -1 and 1.

    • profile image


      16 months ago

      examples given in the article are very nice


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