# Sequences and Series Formulas

In mathematics, sequence and series are the fundamental concepts of arithmetic. A sequence is also referred to as a progression, which is defined as a successive arrangement of numbers in an order according to some specific rules. A series is formed by adding the elements of a sequence. Let us consider an example to understand the concept of a sequence and series better. 1, 3, 5, 7, 9 is a sequence with five terms, while its corresponding series is 1 + 3 + 5 + 7 + 9, whose value is 25. Sequences and series are classified into different types based on the set of rules which are used to form them.

### Sequence and Series Definition

A sequence is defined as a successive arrangement of numbers in an order according to some specific rules. Let x_{1}, x_{2}, x_{3}, x_{4},… be the terms of a sequence, where 1, 2, 3, 4,… represents the term’s position in the given sequence.

- Depending upon the number of terms in a sequence, it is classified into two types, namely a finite sequence and an infinite sequence.
- A series is formed by adding the elements of a sequence.

**If x _{1}, x_{2}, x_{3}, x_{4}, ……. is the given sequence, then its corresponding series is given by S_{N} = x_{1}+x_{2}+x_{3} + .. + x_{N} **

- Depending on whether the sequence is finite or infinite, the series can be either finite or infinite.

**Sequence vs Series**

Sequence | Series |
---|---|

A sequence is defined as a successive arrangement of numbers in an order according to some specific rules. | A series is formed by adding the elements of a sequence. |

It is basically a grouping of components that follow a certain pattern. | It is a sum of elements that follow a pattern. |

In a sequence, the order of the numbers is important. | In a series, the order of numbers is not important. |

A finite arithmetic sequence: 3, 5, 7, 9, 11 An infinite geometric sequence: 2, 4, 8, 16, …….. |
A finite arithmetic series: 3 + 5 + 7 + 9+ 11 An infinite geometric series: 2 + 4 + 8 + 16 + …….. |

### Types of Sequences and Series

Sequences and series are classified into different types. Some of the most commonly used examples of sequences and series are:

- Arithmetic Sequences and Series
- Geometric Sequences and Series
- Harmonic Sequences and Series
- Fibonacci Numbers

**Arithmetic Sequence and Series**

An arithmetic sequence is a sequence where each term of the sequence is formed either by adding or subtracting a common term from the preceding number, and the common term is called the common difference. An arithmetic series is referred to as a series developed by using an arithmetic sequence.

**For example,**

2, 5, 8, 11, 14,… is an arithmetic sequence with a common difference of 3, and 2 + 5 + 8 + 11 + 14 +… is the corresponding arithmetic series.

**Geometric Sequence and Series**

A geometric sequence is a sequence where each term of the sequence is formed either by multiplying or dividing a common term with the preceding number, and the common term is called the common ratio. A geometric series is referred to as a series developed by using a geometric sequence. Depending upon the number of terms in a geometric progression it is classified into two types, namely, finite geometric progression and infinite geometric progression.

**For example,**

1, 5, 25, 125, 625,… is a geometric sequence with a common ratio of 5, and 1 + 5 + 25 + 125 + 625 +… is its corresponding geometric series.

**Harmonic Sequence and Series**

A harmonic sequence is a sequence where each term of the sequence is the reciprocal of the element of an arithmetic sequence. A harmonic series is referred to as a series developed by using a harmonic sequence.

**For example,**

2, 5, 8, 11, 14,… is an arithmetic sequence. Now, the harmonic sequence is 1/2, 1/5, 1/8, 1/11, 1/14,… and its corresponding harmonic series is 1/2 + 1/5 + 1/8 + 1/11 + 1/14 +…

**Fibonacci Numbers**

Fibonacci Numbers are a sequence of numbers where each term of the sequence is formed by adding its preceding two numbers, and the first two terms of the sequence are 0 and 1.

As the first term, F_{0}, and the second term, F_{1} of the Fibonacci sequence are 0 and 1, the third term will be, F_{2} = F_{1} + F_{0 }= 1 + 0 = 1.

Similarly,

The fourth term, F_{3} = F_{2} + F_{1} = 1 + 1 = 2

The fifth term, F_{4} = F_{3} + F_{2} = 2 + 1 = 3

The sixth term, F_{5} = F_{4 }+ F_{3} = 3 + 2 = 5

Therefore, the (n+1)^{th} term of the Fibonacci sequence can be expressed as,** F _{n} = F_{n-1} + F_{n-2}. **

The numbers of a Fibonacci sequence are given as: 0, 1, 1, 2, 3, 5, 8, 13, 21, 38, . . .

### Sequence and Series Formulae

| ## Arithmetic Progression | ## Geometric Progression |
---|---|---|

## Sequence | a, (a + d), (a+2d), (a + 3d),………. | a, ar, ar^{2},ar^{3},…. |

## Series | a + (a + d) + (a + 2d) + (a + 3d) +… | a + ar + ar^{2} + ar^{3} +…. |

## First term | a | a |

## Common Difference or Ratio | Common difference = Successive term – Preceding term => d = a | Common ratio = Successive term/Preceding term => r = ar |

## nth term | a + (n-1)d | ar^{(n-1)} |

## Sum of first n terms | S_{n} = (n/2)[2a + (n-1)d] | S S |

- The sum of the terms of an
**infinite geometric series**is given by,

S_{n}= a/(1−r)for |r| < 1, and

not defined for |r| > 1

**Sample Problems**

**Problem 1: Using the sequence and series formula, determine the seventh term of the given geometric sequence: 3, 1, 1/3, 1/9, 1/27, 1/81, ___.**

**Solution:**

Given sequence: 3, 1, 1/3, 1/9, 1/27, 1/81, ___

Now, a = 3, r = 1/3

By using the formula for the nth term of a geometric sequence and series:

a

_{n}= ar^{(n-1)}Putting the known values in the formula:

a

_{7}= 3 × (1/3)^{(7-1)}a

_{7}= 3 × (1/3)^{6}a

_{7}= (1/3)^{5}= 1/243Hence, the seventh term of the given series is 1/243.

**Problem 2: Using the sequence and series formula, find the 10th term of the arithmetic sequence 14, 10, 6, 2, -2, -6, ___.**

**Solution:**

Given sequence: 14, 10, 6, 2, -2, -6, ___

Now, a = 14

d = 10 -14 = -4

Using the formula for the nth term of an arithmetic sequence:

a

_{n}= a+(n-1)da

_{10}= 14 + (10 – 1)(-4)a

_{10}= 14 + (9)(-4)a

_{10}= 14 – 36 = -22Hence, the 10th term of the sequence is -22.

**Problem 3: If p, q, and r are in A.P., find the value of (q ^{2}-pr)/(p – q)^{2}.**

**Solution:**

Given that p, q, and are in A.P

let p, q, and r be a-d, a, a + d.

So, p = a-d, q = a, r = a + d

p – q = a- d – (a + d) = -2d

(p – q)

^{2}= (-2d)^{2}= 4d^{2}q

^{2}= a^{2}p × r = (a – d) (a + d) = (a

^{2}– d^{2})q

^{2}– pr = a^{2}– (a^{2}– d^{2}) = d2So, (q

^{2}– pr)/(p – q)^{2}= d^{2}/4d^{2}= 1/4Hence, the value (q

^{2}-pr)/(p – q)^{2 }= 1/4.

**Problem 4: Find the sum of the infinite geometric series 1, -2/3, 4/9, -8/27, 16/81___.**

**Solution:**

Given sequence: 1, – 2/3, 4/9, -8/27, 16/81___

Now, a = 1,

The common ration of the sequence, r = (-2/3)/1 = -2/3

By using the sequence and series formulas,

Sum of the given series = a/(1 – r)

= 1/(1 – (-2/3))

= 1/(1 + 2/3)

= 1/(5/3) = 3/5

Hence, the sum of the infinite geometric series is 3/5.

**Problem 5: Determine the sum of the first 15 terms of the sequence 0.5, 0.55, 0.555,___ up to 15 terms.**

**Solution:**

Given sequence: 0.5, 0.55, 0.555,___up to 15 terms

⇒ 0.5 + 0.55 + 0.555 + 0.5555, …….. up to 15 terms

⇒ 5[0.1 + 0.11 + 0.111 + 0.1111, …….. up to 15 terms]

⇒ (5/9)[0.9 + 0.99 + 0.999 + 0.9999, …… up to 15 terms]

⇒ (5/9) [(1 – 0.1) + (1 – 0.01) + (1 – 0.001), …… up to 15 terms]

⇒ (5/9) [(1 + 1 + 1 + 1, ……. up to 15 terms) – (0.1 + 0.01 + 0.001 + 0.0001 + ….. up o 15 terms)]

⇒ (5/9) [15 – (0.1) (1 – (0.1)

^{15})/(1 – 0.1)]⇒ (5/9) [15 – (0.1)(1 – (0.1)

^{15})/(0.9)]⇒ (5/9) [15 – (1/9) {1 – (0.1)

^{15}}] as 1 – (0.1)^{15 }= 1 (approx)⇒ (5/9) (1/9) [134 ]

⇒ 8.27 (approx)

**Problem 6: Determine the nth term of the given series: 2, (2 + 4), (2 + 4 + 6), (2 + 4 + 6 + 8),…..**

**Solution:**

Here by observing the sequence,

nth term = (2 + 4 + 6 + 8 + 10 . . . . . . . . . . . .+ 2n)

The nth term is an arithmetic series in itself with

first term (a) = 2andcommon difference (d) = 2Now,

Sum of n terms of an Arithmetic progression is (n/2)[2a + (n-1)d]

= (n/2)[2 × 2 + (n-1) × 2]

= (n/2) × 2 [ 2 + (n – 1)]

= n(n+1)

Hence,

the nth term of the given series is n(n+1).

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