Sequence
It means an arrangement of number in definite order according to some rule
Example
$2,4,6,8,10, 12,14.........$
$1, 3, 5,9, 11, 13........$
$1^2 +2^2 + 3^3 +..........$
Important point
 The various numbers occurring in a sequence are called its terms. The number of terms is called the length of the series
 Terms of a sequence are denoted generally by $a_1 , a_2, a_3, ....., a_n$ etc., The subscripts denote the position of the term. We can choose any other letter to denote it
 The nth term is the number at the nth position of the sequence and is denoted by a_{n}
 The nth term is also called the general term of the sequence
 A sequence can be regarded as a function whose domain is the set of natural numbers or some subset of it of the type {1, 2, 3...k}.
 Many times is possible to express general term in terms of algebraic formula. But it may not be true in other cases. But we should be able to generate the terms of the sequence using some rules or theoretical scheme
Example
1. 0,2, 4,6,8.......
General Term a_{n} = 2n
2. 1, 1, 2, 3, 5, 8,..
This sequence does not have visible algebraic formula, but the sequence is generated by the recurrence relation given by
a_{1} =a_{2} =1
an =a_{n2} +a_{n1} , n > 2
This sequence is called Fibonacci sequence
Finite sequence
A sequence containing a finite number of terms is called Finite sequence
infinite sequence
A sequence is called infinite if it is not a finite sequence.
Series
Let a
_{1} , a
_{2} , a
_{3} ..be the sequence, then the sum expressed a
_{1} + a
_{2} +a
_{3} + ....... is called series.
 A series is called finite series if it has got finite number of terms
 A series is called infinite series if it has got infinite terms

Series are often represented in compact form, called sigma notation, using the Greek letter σ (sigma)

so , a_{1} + a_{2} +a_{3} + .......a_{n} can be expressed as $\sum_{k=1}^{n}a_k$
Example
$1+2+3 + 4+ 5+....+100$
$= \sum_{k=1}^{n}k$
Types of Sequences
Arithmetic Progression
An arithmetic progression is a sequence of numbers such that the difference of any two successive members is a constant
Examples
1. $1,5,9,13,17....$
2. $1,2,3,4,5,.....$
Important Notes about Arithmetic Progression
1.The difference between any successive members is a constant and it is called the common difference of AP
2. If a
_{1}, a
_{2},a
_{3},a
_{4},a
_{5} are the terms in AP then
D=a
_{2 }a
_{1} =a
_{3}  a
_{2} =a
_{4} – a
_{3}=a
_{5} –a
_{4}
3. We can represent the general form of AP in the form
a,a+d,a+2d,a+3d,a+4d..
Where a is first term and d is the common difference
4. Nth term of Arithmetic Progression is given by n
^{th}term = a + (n  1)d
5. Sum of nth item in Arithmetic Progression is given by
$S_n =\frac {n}{2}[a + (n1)d]$
Or
$S_n =\frac {n}{2}[t_1+ t_n]$
Arithmetic Mean
The arithmetic mean A of any two numbers a and b is given by (a+b)/2 i.e. a, A, b are in AP
A a = b A
or $A = \frac {(a+b)}{2}$
If we want to add n terms between A and B so that result sequence is in AP
Let A
_{1}, A
_{2} , A
_{3} , A
_{4} , A
_{5} .... A
_{n} be the terms added between a and b
Then
b= a+ [(n+2) 1]d
or $d = \frac {(ba)}{(n+1)}$
So terms will be $a+ \frac {(ba)}{(n+1)}, a + 2 \frac {(ba)}{(n+1)} ,....... a+ n \frac {(ba)}{(n+1)}$
Geometric Progressions
A sequence is said to be a geometric progression or G.P., if the ratio of any term to its preceding term is same throughout.
Examples
1. $2,4,8,16,32....$
2. $3,6,12,24,48,.....$
Important Notes about Geometric Progressions

The ratio of any term to its preceding term is same throughout. This constant factor is called the common ratio.

If a_{1}, a_{2},a_{3},a_{4},a_{5} are the terms in AP then
$r=\frac {a_2}{a_1} =\frac {a_3}{a_2} =\frac {a_4}{a_3}=\frac {a_5}{a_4}$

We can represent the general form of G.P in the form
a,ar,ar^{2} ,ar^{3} ......
Where a is first term and r is the common ratio
 Nth term of Geometric Progression is given by n^{th}term = ar^{n1}
 Sum of nth item inGeometric Progression is given by
$S_n =a \frac {r^n 1}{r1}$
Or
$S_n =a \frac {1r^n }{ 1r}$
Geometric Means
The Geometric mean G of any two numbers a and b is given by (ab)
^{1/2} i.e. a, G, b are in G.P
$\frac {G}{a} = \frac {b}{G}$
or $G=\sqrt {ab}$
If we want to add n terms between A and B so that result sequence is in GP
Let G
_{1}, G
_{2} , G
_{3} , G
_{4} , G
_{5} .... G
_{n} be the terms added between a and b
Then
$b= ar^{n+1}$
or $r= (\frac {b}{a})^{\frac {1}{n+1}}$
So terms will be $a(\frac {b}{a})^{\frac {1}{n+1}}, a(\frac {b}{a})^{\frac {2}{n+1}} ,....... a(\frac {b}{a})^{\frac {n}{n+1}}$
Relationship Between A.P and G.P
Let A and G be A.M. and G.M. of two given positive real numbers a and b, respectively.
Then
A ≥ G
And
$A  G = \frac {{\sqrt {a}  \sqrt {b}}^2}{2}$
Sum of Special Series
Sum of first n natural numbers
$1 + 2 + 3 +..... + n = \frac {n(n+1}{2}$
$ \sum_{k=1}^{n} n =\frac {n(n+1}{2}$
sum of squares of the first n natural numbers
$1^2+ 2^2+3^2 +.... + n^2 = \frac {n(n+1)(2n+1)}{6}$
$ \sum_{k=1}^{n} n^2= \frac {n(n+1)(2n+1)}{6}$
Sum of cubes of the first n natural numbers
$1^3+ 2^3+3^3 +..... + n^3 = \frac {(n(n+1))^2}{4}$
$ \sum_{k=1}^{n} n^3= \frac {(n(n+1))^2}{4}$
Rules for finding the sum of Series
a. Write the nth term T
_{n}of the series
b. Write the T
_{n} in the
polynomial form of n
$T_n= a n^3 + bn^2 + cn +d$
c. The sum of series can be written as
$ \sum S = a \sum {n^3} + b \sum {n^2} + c \sum n + nd
We already know the values if these standard from the formula given above and we can easily find the sum of the series
Example
Find the sum of the series
$2^2 + 4^2 + 6^2 + .....(2n)^2}
Solution
Let nth term T
_{n}of the series
Then
$T_n = (2n)^2 = 4n^2$
Now
$2^2 + 4^2 + 6^2 + .....(2n)^2 = \sum_{k=1}^{n} 4k^2 = 4 sum_{k=1}^{n} k^2 = 4 \frac {n(n+1)(2n+1)}{6}$
$=\frac {2n(n+1)(2n+1)}{3}$
Method of Difference
Many times , nth term of the series can be determined. For example
5 + 11 + 19 + 29 + 41......
If the series is such that difference between successive terms are either in A.P or G.P, then we can the nth term using method of difference
$S_n = 5 + 11 + 19 + 29 + ... + a_{n1} + a_n$
or $S_n= 5 + 11 + 19 + ... + + a_{n1} + a_n$
On subtraction, we get
$0 = 5 + [6 + 8 + 10 + 12 + ...(n – 1) terms] – a_n$
Here 6,8,10 is in A.P,So
$a_n = 5 + \frac {n1}{2} [12 + (n1)2]$
or $ a_n= n^2 + 3n + 1$
Now it is easy to find the Sum of the series
$S_n = \sum_{k=1}^{n} {k^2 +3k +1}$
$=\frac {n(n+2)(n+4)}{3}$
Also Read
 Notes
 NCERT Solutions & Worksheets
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