# Sequence and series

We often come across the word “ sequence and series” in our everyday life. If you notice most of the things around us happens in sequence and series.
Like the days of the week Monday comes after Sunday, Tuesday comes after Monday and so on. Thus, the days in week occur in sequence.In school morning assembly teachers tells the students to stand according to your height.

Let us understand the sequence and series mathematically.

SEQUENCE A sequence can be defined as a proper arrangement of elements in a particular order.Each elements in the sequence are separated by “comma(,)”.
For example:
1, 2, 3, 4, … are in sequence. Here the pattern is – “Each element is +1 greater than the previous element”.

SERIES The sum of the elements of the sequence is called a series.
To elaborate, let us take an example from above.
We know, 1, 2, 3, 4, … are in sequence.
If we add all the number of the above sequence,
1+2+3+4+… Then this is called a series.

In the Sequence and Series topic we will learn the following

• Arithmetic Progression
• Geometric Progression
• Harmonic Progression

### ARITHMETIC PROGRESSION

Arithmetic Progression Definition

Arithmetic Progression(A.P.) – In an arithmetic progression, the difference between the two consecutive terms is always the same.
In other words, It is a sequence in which each term increases or decreases by a fixed constant.

The fixed difference between the consecutive terms is called the “common difference” of the Arithmetic Progression.It is denoted by “d”.
It is to be noted, that this difference can be positive, negative, or zero.

Each number in the Arithmetic Progression is called a term.

In an Arithmetic Progression, if “a” is the first term and “d” is the common difference, Then
the general form of an arithmetic progression is given by,
a, a+d, a+2d, a+3d,…..

Also,If { a }_{ 1 }, { a }_{ 2 }, { a }_{ 3 },…. { a }_{ n } are in AP then common difference d is given by,

d= { a }_{ 2 }{ a }_{ 1 } = { a }_{ 3 }{ a }_{ 2 } = { a }_{ 4 }{ a }_{ 3 }=……= { a }_{ n }{ a }_{ n-1 }

When there are definite number of terms in an arithmetic progression then it is called finite arithmetic progression.
Moreover,  if there are indefinite number of terms in an arithmetic progression then it is called infinite arithmetic progression.

nth term of an Arithmetic Progression:

When in a finite arithmetic series-

“a” is the first term
“d” is the common difference
“l” is the last term
{ a }_{ n } is the nth term of the AP

Thus,

 nth term of the Arithmetic Progression is given by-                          { a }_{ n }  = a + (n-1) d

Also,

1. Three consecutive terms in an A.P is given by-
a – d, a , a+ d

2. Four consecutive terms in an A.P is given by-
a -3d, a-d, a+d, a+3d

Sum of n terms In an AP

The sum of first n term in an A.P. whose first term is “a” and common difference is “d” is given by-

#### { S }_{ n } = \frac { n } { 2 } [2a + (n-1)d]

Also , if “l” is the last term of A.P
then,

#### { S }_{ n } = \frac { n } { 2 } [a + l ]

⇒ The nth term of an AP is the difference of the sum to first n term and the sum to first (n-1) terms  of it.
i.e   { a }_{ n } ={ S }_{ n }{ S }_{ n-1 }

To summarise, Arithmetic Mean

Suppose there are two numbers a and b. We want to insert a number “A” between them such that a,A,b are in A.P. Here A is the arithmetic mean of the numbers a and b.
Also,
as a, A, b are in A.P.
⇒    A – a = b – A  {as the difference of consecutive numbers in AP are equal}

#### A = \frac { a+b }{ 2 }

Now, We can also insert n numbers { A }_{ 1 }, { A }_{ 2 }, { A }_{ 3 },…. { A }_{ n } between a and b. Such that a,{ A }_{ 1 }, { A }_{ 2 }, { A }_{ 3 },…. { A }_{ n } ,b are in A.P.

As a,{ A }_{ 1 }, { A }_{ 2 }, { A }_{ 3 },…. { A }_{ n } ,b are in AP.
Here the total number of terms is (n+2).
and b is the (n+2)th term.
So,
b = a +[(n+2)-1] d
⇒   b = a + (n + 1)d
Thus,

#### d=\frac { b-a }{ n+1 }

Thus common difference of this AP is d=\frac { b-a }{ n+1 }

and the terms are
⇒here { A }_{ 1 } =a + d = a+ (\frac { b-a }{ n+1 })

{ A }_{ 2 } =a + 2d = a+ 2(\frac { b-a }{ n+1 })
.
.
.
{ A }_{ n } =a + nd = a+ n(\frac { b-a }{ n+1 })

### GEOMETRIC PROGRESSION

GEOMETRIC PROGRESSION DEFINITION

Geometric Progression– A series of non zero numbers, in which every number after the first number can be found by multiplying its immediately preceding number by a constant.

In short,
A sequence { a }_{ 1 }, { a }_{ 2 }, { a }_{ 3 },…, { a }_{ n } is called a geometrical progression if each term is non zero and

###### \frac { { a }_{ k+1 } }{ { a }_{ k } } = r(constant) for k ≥ 1.
nth term of a Geometric Progression

#### nth term of a Geometric Progression

If “a” is the first term and “r” is the common ratio
Then,
nth term of a G.P is given by-

#### A G.P can be infinite or finite.

• Finite Geometric progression can be written as-
a, ar, a{ r }^{ 2 }, a{ r }^{ 3 }, . . .,a{ r }^{ n-1 }
• Infinite Geometric Progression can be written as-
a, ar, a{ r }^{ 2 }, a{ r }^{ 3 }, . . .,a{ r }^{ n-1 },…
Sum to n terms of a Geometric Progression

#### Sum to n terms of a Geometric Progression

If “a” is the first term and “r” is the common ratio
then, sum of n terms is given by

{ S }_{ n }=a + ar + a{ r }^{ 2 } + a{ r }^{ 3 } + . . .+ a{ r }^{ n-1 }

The sum can be found by three cases depending on the common ratio”r”.

1. If  r = 1
then
 { S }_{ n }= na
2. If r < 1
then
 { S }_{ n }=a(\frac { 1-{ r }^{ n } }{ 1-r } )
3. If r> 1
then
 { S }_{ n }=a(\frac { { r }^{ n }-1 }{ r-1 } )

#### The sum of Infinite G.P is for -1< r< 1 or |r|< 1

##### { S }_{ ∞ }=\frac { a }{ 1-r }
Geometric Mean

Suppose there are two positive integers a and b. We want to insert “b” between a and b, such that a, c, b are in G.P. Then c=\sqrt { ab }  and is called the Geometric Mean.

We can insert many numbers between a and b, so that the resulting series will be in G.P.
Let, { G }_{ 1 }, { G }_{ 2 }, { G }_{ 3 }, { G }_{ 4 },…,{ G }_{ n } be n numbers between a and b, such that
a,{ G }_{ 1 }, { G }_{ 2 },{ G }_{ 3 },…,{ G }_{ n }, b are in G.P.

Here the total number of terms = (n+2)
and “b” is the (n+2)th term
Then,
b = a{ r }^{ [(n+2)-1] }
b = a{ r }^{ n-1 }
⇒         r ={ (\frac { b }{ a } })^{ \frac { 1 }{ n+1 } }

##### r ={ (\frac { b }{ a } })^{ \frac { 1 }{ n+1 } }

Hence,

The { G }_{ 1 }= ar =a{ (\frac { b }{ a } })^{ \frac { 1 }{ n+1 } }

and { G }_{ 2 }= a{ r }^{ 2 }=a{ (\frac { b }{ a } })^{ \frac { 2 }{ n+1 } }.
.
,

##### { G }_{ n }= a{ r }^{ n }={ (\frac { b }{ a } })^{ \frac { n }{ n+1 } }

To summarise, ### Harmonic Progression

Harmonic Progression Definition

Harmonic progression can be defined as a sequence of non zero numbers in which the sequence of  their reciprocal forms an Arithmetic progression.
In simple words, The sequence a,b,c,d are said to be in harmonic progression when \frac { 1 }{ a }, \frac { 1 }{ b },\frac { 1 }{ c },\frac { 1 }{ d } are in Arithmetic progression.

For example,
\frac { 1 }{ 3 }, \frac { 1 }{ 5 },\frac { 1 }{ 7 },\frac { 1 }{ 7 } are in harmonic progression because their reciprocal are 3,5,7,9 are in Arithmetic Progression with common difference 2.

nth term of an Harmonic Progression

if “a” is the first term
and “d” is the difference
then, the n th term is

{ a }_{ n }=\frac { 1 }{ [a+(n-1)d] }

If you notice , the nth term of an HP is the reciprocal of nth term of an AP.

Harmonic Mean
We can insert a number “H” between a and b .Such that a,H,b are in HP.
Here “H” is called the Harmonic mean.It is given by,
H = \frac { 2ab }{ a+b }

### Relationship between Arithmetic mean, Geometric mean and harmonic mean

We know that for two numbers “a” and “b” ,

ARITHMETIC MEAN =\frac { a+b }{ 2 }

GEOMETRIC MEAN =\sqrt { ab }

HARMONIC MEAN = \frac { 2ab }{ a+b }

Then,

 \frac { a+b }{ 2 }  ≥ \sqrt { ab }≥ \frac { 2ab }{ a+b }

i.e.

 AM ≥GM ≥HM

Also,
AM × HM = \frac { a+b }{ 2 }× \frac { 2ab }{ a+b }
⇒AM× HM    = ab       ………….eq (i)
and
GM = \sqrt { ab }
{ GM }^{ 2 } = ab   ……………..eq(ii)
Thus from equation (i) and (ii)
we can say
{ GM }^{ 2 }=AM × HM

 GM = \sqrt { AM×HM }

### Sum to n terms of some special series

• 1 + 2 + 3 + 4 + 5 + … + n =\frac { n(n+1) }{ 2 }

• { 1 }^{ 2 }+{ 2 }^{ 2 }+{ 3 }^{ 2 }+{ 4 }^{ 2 }+…+{ n }^{ 2 }=\frac { n(n+1)(2n+1) }{ 2 }

• { 1 }^{ 3 }+{ 2 }^{ 3 }+...+{ n }^{ 3 }={ [\frac { n(n+1) }{ 2 } ] }^{ 2 }

It is recommended that you must understand sequence and series thoroughly. As its concepts play a huge role in the arithmetic part of mathematics.

Other important topics you might want to look at-