# Sequence and Series - Important concepts, Formulas & Tricks

Most of the logically puzzled questions asked from Sequences and Progressions. An aspirant's approach towards this topic should be more logical, at the same time he/she should highly aware about the formula and results. The aptitude questions on progressions will be direct type of progression format or any other type of sequences. This article explain in detail different types of sequence and series along with important concepts, formulas and tricks to solve the aptitude problems easily.

The first and important logical responsibility from the student's end is to identify the nature of the sequences which is directly or indirectly given in the questions. With your logical reasoning skill and by practice you can crack the first half of this requirement. Then things are remaining with application of results and mere calculation. In each and every placement test and competitive exam, you can expect one or two questions from Sequences and Progressions.

The articles covers following concepts
• Sequence
• Series
• Progression
• Arithmetic Progression (AP)
• Properties of AP
• Arithmetic Mean (AM)
• Insertion of AM between two numbers
• Geometric Progression (GP)
• Infinite GP
• Properties of GP
• Geometric Mean (GM)
• Insertion of GM between two numbers
• Arithmetico - Geometric Progressions (AG)
• Harmonic Progression (HP)
• Harmonic Mean
• Properties of AM, GM and HM.

## Understanding basic terminologies.

Sequence:

If a1, a2, a3, .........an is an arrangement of 'n' numbers (more often non- zero numbers) as per a well defined order is called a sequence, and it is denoted by {an}. Here the numbers a1, a2, a3, .........an are called the elements or terms of the sequence.

If there is finite number of terms in a sequence, then it is called a 'finite sequence'. If there is infinite number of terms then the sequence is called an infinite sequence.

Eg: 1/3, 1/6, 1/9 ......... is a sequence.

Series:

If a1, a2, a3, .........an is a sequence of 'n' terms then their sum a1+ a2+ a3+......... + an is called a finite series and it is denoted by ∑n.

Eg: 1, 2, 3, 4,........., 50 is a finite sequence of first 50 natural numbers, then 1 + 2 + 3 + 4 + .....+ 50 is called the series corresponding to the given sequence.

## Progressions:

Special types of sequences are called progressions. Progressions are mainly classified in to three.

• Arithmetic Progressions (AP)
• Geometric Progressions (GP)
• Harmonic Progressions (HP)

### Arithmetic Progressions (AP)

A sequence a1, a2, a3, .........an is called an AP, when a2 - a1 = a3 - a2 = ....... = an - an-1.

ie. The terms of an AP is successively increase or decrease by a constant value and this constant value of difference is called the 'common difference' of the AP, denoted as 'd'.

If a1, a2, a3, .........an is an AP, then d = an - an-1

d = an - an-1

Example:

 Arithmetic Progression d 5, 7, 9, 11, ..... 2 10, 20, 30, 40, 10 99, 98, 97, 96,........ -1 1/2, 1, 3/2, 2, .... 1/2

Important results of Arithmetic Progression:

Let a1, a2, a3, .........an is an AP of 'n' terms

• Common difference (d) = an - an-1
• nth term (an) = a + (n - 1)d
• Number of terms (n) = [(an - a1)/ d] + 1
• Sum of 'n' terms = n/2 [2a + (n - 1)d] OR n/2 [a1 + an]
• Arithmetic Mean (AM) or Average of n terms of an AP = (a1 + an) / 2
ie. AM = (a1 + an) / 2 = (a2 + an-1) / 2 = (a3 + an-2) / 2
• In an Arithmetic Progression, Mean = Median (where median is the middlemost term in the arrangement)

### Properties of Arithmetic Progression

• If a1, a2, a3, .........an is an AP with common difference 'd', then a1 ± k, a2 ± k, a3 ± k, .........an ± k is also an AP with the same common difference 'd', where 'k' is any real constant.
• If a1, a2, a3, .........an is an AP with common difference 'd', then ka1, ka2 , ka3, ........., kan is also an AP with common difference 'kd', where 'k' is any real constant.
• In an AP the sum of any two terms which are equidistant from both the ends are equal.
ie. If a1, a2, a3, .........an be an AP then,
a1 + an = a2 + an-1 = a3 + an-2 = . = am + an-m+1
• If the algebraic expression of an AP is Pn + Q, where P and Q are any two real constants, then P (the coefficient of n) is the common difference of the AP and the first term of the AP is 'P + Q'.
• If the algebraic expression in the form Pn2 + Qn, where P and Q are any two real constants, is representing the sum to 'n' terms of an AP, then the common difference of this AP is '2P' and the first term is 'P + Q'.
• Three numbers a, b, c are three consecutive terms of an AP if 2b = a + c.
• The sequence formed by the selected terms of an AP at regular intervals will be an AP.

### Insertion of Arithmetic Means between two numbers

Two quantities 'a' and 'b' are given. If we have to insert 'n' quantities A1, A2, A3, ........., An in between 'a' and 'b', such that a, A1, A2, A3, ...., An, b form an AP, then A1, A2, A3, ........., An are the Arithmetic Means in between 'a' and 'b'.

Illustration:

Insert 'n' AM's between 'a' and 'b'.

Let A1, A2, A3, ........., An are the 'n' AM's between 'a' and 'b'.

Then a, A1, A2, A3, ....., An, b is an AP of 'n + 2' terms.

(n+2)th term = a + (n + 1) d
a + (n + 1) d= b
d = (b-a) / (n+1)
A1 = a + d = a + (b-a) / (n+1)
A2 = a + 2d = a + 2 * [(b-a) / (n+1)]
A3 = a + 3d = a + 3 * [(b-a) / (n+1)]
.
.
.
An = a + nd = a + n * [(b-a) / (n+1)]
 In general, the mth AM among the 'n' AM's between 'a' and 'b' can be found by Am = a + m * [(b-a) / (n+1)]

Example:

Insert four Arithmetic Means between 5 and 40.

Solution:

n = 4
a = 5
b = 40
A1 = 5 + 1 * [(40-5) / (4+1)] = 12
A2 = 5 + 2 * [(40-5) / (4+1)] = 19
A3 = 5 + 3 * [(40-5) / (4+1)] = 26
A4 = 5 + 4 * [(40-5) / (4+1)] = 33
Required AM's are 12, 19, 26 and 33.

Insertion of a single AM between 'a' and 'b'

Let 'x' be the single AM between two numbers 'a' and 'b', then x = (a + b)/2

For better understanding of the Arithmetic Progression, let us look at an example.

Two numbers have a sum of. A certain number of Arithmetic Means are being inserted between them and the sum of these newly inserted AM's is 3 more than the number of AM's. How many AM's inserted?

A)54  B)62  C)31  D)26

Solution

Let A1, A2, A3,..., An are the 'n' AM's.
A1+ A2+ A3+ .....+ An = n/2 * [A1 + An]
= n/2 * (19/9)
= 19n/18
It is given that, A1+ A2+ A3+ ......+ An = n + 3
ie. 19n/18 = n+3
19n/18 - n =3
1/18 * n = 3
n=54
ie. There are 54 new AM's inserted.

### Geometric Progression (GP)

A sequence of non-zero numbers a1, a2, a3, .........an is called a GP, if the ratio of term and its preceding term is always a constant.

ie. a2 / a1 = a3 / a2 = .... = an / an-1

This constant ratio is called the 'Common ratio' of the GP and denoted as 'r'.

ie. r = an / an-1

Eg: 2, 4, 8, 16,..... is a GP

r = 4/2 = 8/4 = 2

Eg: 1/3, 1/9, 1/27, 1/81,...... is a GP.

r = (1/9) / (1/3) = (1/3)

### Important results of Geometric Progression

#### nth term of a GP

If the first term of a GP is 'a' and its common ratio is 'r', then the GP is,

a, ar, ar2, ar3,......, arn-1

ie. The nth term = arn-1

 nth term = arn-1, where 'a' is the first term and 'r' is the common ratio

Eg: Find the 11th term of the GP 3, 6, 12, 24, ....

a = 3 r = 2 a11 = ar11-1 = 3 * 210 = 3 * 1024 = 3072

#### Sum to 'n' terms of a GP

Let Sn denote the sum to 'n' terms of a GP with first term 'a' and common ratio 'r'.

Sn = [(rn - 1) / (r - 1)], where r > 1
OR
Sn = [(1 - rn) / (1 - r)], where r < 1
If the last term of GP is l, then l = arn-1
Sn = a[(rn - 1) / (r - 1)]
= arn * (r - a) / (r - 1)
= (lr - a)/(r - 1) where r > 1
OR
Sn = (a - lr) / (1 - r) where r < 1

Eg: Find the sum to 8 terms of the GP: 5, 10, 20, 40,......

a = 5
r = 2 [ie. r > 1]
Sn = a[(rn - 1) / (r - 1)]
= 5[(28 - 1) / (2 - 1)]
= 5 * 255
= 1275

Eg: Find the sum to 5 terms of the GP: 3, 3/2, 3/4, 3/8, ...

Solution:

a = 3
r = 1/2 [ie. r < 1]
Sn = a * [(1 - rn) / (1 - r)]
= 3 * [(1 - 1/25) / (1 - 1/2)]
= 3 * [(1 - 1/32) / (1 - 1/2)]
= 3 * 31/32 = 93/16

#### Sum of terms in an infinite GP

If there are infinite terms in a GP, then it is called an infinite GP.

If the common ratio of an infinite GP is lie in between -1 and 1, ie. -1 < r < 1 or |r| < 1, then we can find out the sum of the terms of the GP.

S = a/(1 - r)

Where 'a' is the first term and r is the common ratio.

Eg: Find the sum of the infinite series 4 + 2 + 1 + 1/2 + 1/4 + ......

a = 4
r = 1/2
S = a/(1 - r)
=4/(1 - 1/2)
= 8

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