# Arithmetic Sequences – Sequences and Series | Class 11 Maths

Sequence is an enumerated collection of objects in which repetition are allowed and order matters they formed such a pattern by which we can identify whole series. We can generalize this whole series which is called sequence.

**Example.1. Sequence of even number having difference 4.**

**Solution:**

2, 6, 10, 14,……………..a

_{n}.Here in the above example the first term of sequence is a

_{1}=2. And last term is a_{n.}

**Example 2. An arrangement of numbers such as 1, 1, 2, 3, 5, 8,… Has no visible, pattern, but the sequence is generated by the recurrence relation given by.**

**Solution:**

a

_{1}=1, a_{2}=1, a_{3}=2a

_{3}=a1+a2.a

_{n}=a_{n-2}+a_{n-1 }Where n>2This is called Fibonacci sequence.

**Example.3.An arrangement of numbers say 2,8,14,20,… has a visible pattern and its sequence is generated by the relation.**

**Solution:**

a

_{1}=2, a_{2}=6, a_{3}=14. Common difference between element is constant and is equal to d =a_{2}-a_{1 or}take any two adjacent numbers .So sequence becomes a

_{n}=a_{1}+(n-1)*d.

**Series: **A series can be highly generalized as the sum of all the terms in a sequence . Let a sequence is given as a_{1}, a_{2}, a_{3,} . . . . . . ,a_{n.} Then the expression is called series associated with given sequence.The dependency of finite or infinite series depends on the nature of sequence whether it is finite or infinite.Series are denoted by ∑ (sigma) notation. Thus, the series a1+ a2 +a3+an = ∑^{n}_{k=1 }a_{k.}

## Arithmetic Sequence

In the arithmetic sequence the absolute difference between one term and next term is constant.

Explanation:

A sequence a_{1},a_{2}, … a_{n. Is} Called arithmetic sequence or arithmetic progression if a_{n+1 }– a_{n}=d where d is constant. And it is common difference.

Let’s make an Arithmetic Progression with first term A and common difference D.

{A, A+D, A+2D, A+3D,… ….}

The n^{th }General Term of an A.P is given by a_{n}=a+(n-1)*d.

Where, a is first term of A.P, d is common difference and n is the number of terms .

## Explicit Formula’s

Writing explicit formulas

Let’s take a sequence 6, 16, 26, 36 …76.

The first term of the sequence is 6 and common difference is 10.

We can get any term in the sequence by taking the first term 6 and common difference is 10.

**Calculation for nth term**.

1. | 6 | 6+0.10=6 |

2. | 6+10 | 6+1.10=16 |

3. | 6+10+10 | 6+2.10=20 |

4. | 6+10+10+10 | 6+3.10=36 |

5. | 6+10+10+10+10 | 6+4.10=46 |

6. | 6+10+10+10+10+10 | 6+5.10=56 |

7. | 6+10+10+10+10+10+10 | 6+6.10=66 |

The nth term can be find out easily. The first terms is 6, and we get the difference is 10 in each step.

Above statement can be generalized as 6+(n-1)*d.

In general, this is the standard explicit formula of an arithmetic sequence whose first term is, A, end and common difference is D.

A_{n}= A+(n-1)*D.

**Important Properties Of Arithmetic progression **

- If a constant is added to each term of A.P. then the resulting sequence is also an A.P.
- If a constant is subtracted to each term of A.P. then the resulting sequence is also an A.P.
- If each term of an A.P is multiplied by a constant number. Then the resulting sequence is also an A.P.
- If each term of an A.P is divided by non-zero constant number, then the resulting sequence is also an A.P.

**Sum of arithmetic Progression**

Let’s sequence is given as a, a+d, a+2d, a+3d, ….. a+(n-1)*d.

S_{n }= (n/2)(a + l)

where,

a is the first term

l is the last term of the series and

n is the number of terms in the series

Replacing the last term l by the n^{th}term in equation 3 we get,

n^{th}term = a + (n – 1)d

S_{n }= (n/2)(a + a + (n – 1)d)

S_{n }= (n/2)(2a + (n – 1) x d)

### Example.1. Write first three terms in each of the following sequence defined by

**(i) A _{n}=5n+2(n-1)**

**Solution :**

Solve: (i) put n=1, we get a

_{1}=5.1 + 2(1-1) = 5+ 0 =5Put n=2, we get a

_{2}=5.2+2(2-1) =10+2 =12Put n=3, we get a

_{3}=5.3 + 2(3-1) =15 + 4 =19So first three terms are 5,12, 19.

**(ii) A _{n}=2n +4(n-2) **

**Solution :**

Put n=1, we get a

_{1}=2.1+4(1-2) =2-4 = -2Put n=2, we get a

_{2}= 2.2+4(2-2) =4+ 0 =4Put n=3,we get a

_{3}= 2.3 + 4(3-2) =6+4 =10So the first three terms are -2, 4, 10.

**Example.2. Find the 20**^{th }Term of the given expression.

^{th }Term of the given expression.

**Solution :**

A

_{n}=(n-1)(2-n)(3+n)Solve: put n=20 in given expression,

a

_{20}=(20-1)(2-20)(20+3) = 19*-18·23 = -7886.

NOTE: if question asked find nth term of arithmetic sequence then consider the sequence to be a_{n}=a+(n-1)*d.

### Example.3 Find the sum of all natural numbers lying between 100 and 1000 inclusive which are multiples of 5.

**Solution :**

Solve: first term to be 100 and last terms is 1000 and common difference is 5.

So our formula is S

_{n}=(n/2)[2a+(n-1)*d] .We need to find the number of terms , so number of terms is given by (1000-100)/5 = 900/5 =180.

S

_{180}=(180/2)[2·100 +(180-1)*5].S

_{180}=90*[200+179·5]S

_{180}=90·1095 = 98,550.