Sequence and Series Class 11 Notes 

Sequence : 

A set of numbers arranged in a definite order according to some rule is called a sequence.

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A sequence is a function of natural numbers with codomain is the set of Real Numbers (Complex Numbers) . If Range is a subset of Real Numers (Complex Numbers) then it is called a real sequence (Complex Sequence).

OR

A mapping f : N → A then f (n) = t_n , n ∈ N is called a sequence to be denoted by 

Range < f_n > = { f (1) , f (2) , f (3) , ———- } = { t₁ , t₂ , t₃ , ——- } OR < f_n >  or  { t_n } 

OR

A sequence is a function whose domain is the of natural numbers (N) or some subsets of the type {1 , 2 , 3 , 4 , 5 , ——– , k} . 

Example : 2 , 4, 6 , 8 ,—— is a sequence .

Note : Sequence is said to be finite or infinite sequence according it has finite or infinite number of terms.

Series : 

If  a₁ , a₂ , ——– , a_n , ——– is a sequence , then the expression a₁ + a₂ + —— + a_n + —— is called the series . The series is said to be finite or infinite according as the given series is finite or infinite.

Example : 

(i) 2 + 4 + 6 + 8 + ——- + 20 is a finite series.

(ii) 1 + 3 + 5 + ——– is an infinite series.

Progression :

It is not necessary that the terms of a sequence always follow a certain pattern or they are described by some explicity formula for the nth term. Those sequence whose terms follow certaion patterns are called progression.

OR

It the terms of a sequence are written under specific conditions , then the sequence is called progression.

Arithmetic Progression (AP) :

A sequence is said to be an arithmetic progression , if the difference of a term and the term preceding to it is always same.

The constant difference , generally denoted by d , is called the common difference.

An arithmetic progression (AP) is a sequence whose terms increase or decrease by a fixed number. This fixed number is called the common difference of the A.P. , generally denoted by d.

In other word , 

If  a₁ , a₂ , a₃ , ——- , a_n are in A.P. then a₂ – a₁ = a₃ – a₂ = ——- = a_n  – a_{n -1} = d (say)

If a is the first term and d is the common difference , then A.P. can be written as

a , a + d , a + 2 d , ———- , { a + (n – 1) d } , ——- .

Example : 

(i) 1 , 3 , 5 , 7 , ——– 

(ii) 2 , 4 , 6 , ———

The nth Term of an A.P. :

Let a be the first term , d be the common difference and l be the last term of an A.P. then nth term is given by 

T_n = a + (n -1) d  where  d = T_n – T_{n -1} 

The nth term from last is  T ‘_n = l – (n -1) d

Example : How many terms are there in the A.P.  20 , 25 , 30 , —— , 100 ?

(1) 15                             

(2) 17                           

(3) 12                             

(4) 14

Solution : (2) Let  the number of terms be n.

Given , T_n = 100  , a = 20 , d = 5 

We know ,

$$
\begin{aligned}
& T_n=a+(n-1) d \\
& 100=20+(n-1) 5 \\
& 100-20=(n-1) 5 \\
& n-1=\frac{80}{5} \\
& n-1=16 \\
& n=17
\end{aligned}
$$

∴   Number of terms = 17 

The sum of n Terms of an A.P. : 

Suppose there are n terms of a sequence , whose first term is a , common difference is d and last term is l , then sum of n terms is given by 

$$
\begin{aligned}
S_n & =\frac{n}{2}[2 \mathrm{a}+(\mathrm{n}-1) \mathrm{d}] \\
& =\frac{n}{2}[\mathrm{a}+\mathrm{l}] \end{aligned}
$$

i.e. S_n is called as first n term of an A.P.

Example : The sum of all natural numbers lying between 100 and 1000 , which are multiples of 5 is 

(1) 98400                             

(2) 98450                             

(3) 98436                           

(4) 98455

Solution : (2) The numbers are 105 , 110 , 115  ,  ——– , 995.

Here , first term (a) = 105 & common difference (d) = 110 – 105 = 5

Now , nth term ,

$$
\begin{aligned}
& T_n=a+(n-1) d \\
& 995=105+(n-1) 5 \\
& 995-105=(n-1) 5 \\
& n-1=\frac{890}{5} \\
& n-1=178 \\
& n=179
\end{aligned}
$$

Now, the required sum,
$$
\begin{aligned}
S_n & =\frac{\mathrm{n}}{2}[a+l] \\
& =\frac{179}{2}[105+995] \\
& =\frac{179}{2} \times 1100 \\
& =179 \times 550 \\
& =98450
\end{aligned}
$$

Important Results Related to A.P. : 

• For any sequence : T_n = S_n  – S_{n – 1} 
• For an A.P. : d = S_n – 2 S_{n – 1} + S_{n – 2}
• If  ( a₁ , a₂ , a₃ , ——- , a_n )  → A.P. then ( a₁ ± k  , a₂ ± k , a₃ ± k , ——- , a_n ± k ) → A.P.
• If  ( a₁ , a₂ , a₃ , ——- , a_n )  → A.P. then ( k a₁ , k a₂ , k a₃ , ——- , k a_n ) → A.P.
• If a , b , c are in A.P. ⇒ 2 b = a + c 
• 3 terms in A.P. ⇒ a – d , a , a + d
• 4 terms in A.P. ⇒ a – 3 d , a – d , a + d , a + 3 d 
• 5 terms in A.P. ⇒ a – 2 d , a – d , a , a + d , a + 2 d 
• Sum of terms of an A.P. equidistance from starting and end is equal.

i.e. If  a₁ , a₂ , a₃ , ——- a₄₂ , a₄₃  → A.P. then  a₁ + a₄₃ = a₂ + a₄₂ = a₃ + a₄₁ = ——-

Also Read 

Example : If a₁ , a₂ , a₃ , —– , a_n are in A.P. and a₁ + a₂ + a₃ + ——- + a_n = 114 then a₁ + a₆ + a₁₁ + a₁₆ is equal to :

(a) 98                                     

(b) 38                                           

(c) 64                                           

(d) 76

Solution : (d)  a₁ + a₂ + a₃ + ——- + a_n = 114 then 

a₁ + a₁₆ = x ( Let )

a₄ + a₁₃ = x

a₇ + a₁₀ = x  then 

x + x + x = 114 

3 x = 114 

x = 38 

∴ a₁ + a₆ + a₁₁ + a₁₆

 = (a₁ + a₁₆ ) + (a₆ + a₁₁ ) 

= x + x 

= 2 x 

= 2 × 38

= 76

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