**Sequence and Series Class 11 Notes **

**Sequence :Â **

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

**Join the Group :**

OR

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 â†’ A then f (n) = t_{n} , n âˆˆ N is called a sequence to be denoted byÂ

Range < f_{n} > = { f (1) , f (2) , f (3) , ———- } = { t_{1} , t_{2} , t_{3} , ——- } OR < f_{n} >Â orÂ { t_{n} }Â

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} .Â

**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 :Â **

IfÂ a_{1} , a_{2} , ——– , a_{n} , ——– is a sequence , then the expression a_{1} + a_{2} + —— + 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 :Â **

(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 ,Â

IfÂ a_{1} , a_{2} , a_{3} , ——- , a_{n} are in A.P. then a_{2} – a_{1} = a_{3} – a_{2} = ——- = a_{n}Â – 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 :Â **

(i) 1 , 3 , 5 , 7 , ——–Â

(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 byÂ

**T _{n} = a + (n -1) dÂ **whereÂ d = T

_{n}– T

_{n -1Â }

The nth term from last isÂ T ‘_{n} = l – (n -1) d

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

(1) 15Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â

(2) 17Â Â Â Â Â Â Â Â Â Â Â Â Â Â

(3) 12Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â

(4) 14

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

Given , T_{n} = 100Â , a = 20 , d = 5Â

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}

$$

âˆ´Â Â Number of terms = 17Â

**The sum of n Terms of an A.P. :Â **

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 byÂ

$$

\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 isÂ

(1) 98400Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â

(2) 98450Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â

(3) 98436Â Â Â Â Â Â Â Â Â Â Â Â Â Â

(4) 98455

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

- For any sequence : T
_{n}= S_{n}Â – S_{n – 1}Â - For an A.P. : d = S
_{n}– 2 S_{n – 1}+ S_{n – 2} - IfÂ ( a
_{1}, a_{2}, a_{3}, ——- , a_{n})Â â†’ A.P. then ( a_{1}Â± kÂ , a_{2}Â± k , a_{3}Â± k , ——- , a_{n}Â± k ) â†’ A.P. - IfÂ ( a
_{1}, a_{2}, a_{3}, ——- , a_{n})Â â†’ A.P. then ( k a_{1}, k a_{2}, k a_{3}, ——- , k a_{n}Â ) â†’ A.P. - If a , b , c are in A.P. â‡’ 2 b = a + cÂ
- 3 terms in A.P. â‡’ a – d , a , a + d
- 4 terms in A.P. â‡’ a – 3 d , a – d , a + d , a + 3 dÂ
- 5 terms in A.P. â‡’ a – 2 d , a – d , a , a + d , a + 2 dÂ
- Sum of terms of an A.P. equidistance from starting and end is equal.

i.e. IfÂ a_{1} , a_{2} , a_{3} , ——- a_{42} , a_{43}Â â†’ A.P. thenÂ a_{1} + a_{43} = a_{2} + a_{42} = a_{3} + a_{41} = ——-

**Example :** If a_{1} , a_{2} , a_{3} , —– , a_{n} are in A.P. and a_{1} + a_{2} + a_{3} + ——- + a_{n} = 114 then a_{1} + a_{6} + a_{11} + a_{16} is equal to :

(a) 98Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â

(b) 38Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â

(c) 64Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â

(d) 76

**Solution : (d)Â ** a_{1} + a_{2} + a_{3} + ——- + a_{n} = 114 thenÂ

a_{1} + a_{16} = x ( Let )

a_{4} + a_{13} = x

a_{7} + a_{10} = xÂ thenÂ

x + x + x = 114Â

3 x = 114Â

x = 38Â

âˆ´ a_{1} + a_{6} + a_{11} + a_{16}

Â = (a_{1} + a_{16} ) + (a_{6} + a_{11} )Â

= x + xÂ

= 2 xÂ

= 2 Ã— 38

= 76

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