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# Sequences and Series Formulas

• Last Updated : 14 Jul, 2022

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 x1, x2, x3, x4,… 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 x1, x2, x3, x4, ……. is the given sequence, then its corresponding series is given by  SN = x1+x2+x3 + .. + xN

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

Sequence vs Series

### 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, F0, and the second term, F1 of the Fibonacci sequence are 0 and 1, the third term will be, F2 = F1 + F0 = 1 + 0 = 1.

Similarly,

The fourth term, F3 = F2 + F1 = 1 + 1 = 2

The fifth term, F4 = F3 + F2 =  2 + 1 = 3

The sixth term, F5 = F4 + F3 = 3 + 2 = 5

Therefore, the (n+1)th term of the Fibonacci sequence can be expressed as, Fn = Fn-1 + Fn-2

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

### Sequence and Series Formulae

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

Sn = 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:

an = ar(n-1)

Putting the known values in the formula:

a7 = 3 × (1/3)(7-1)

a7 = 3 × (1/3)6

a7 = (1/3)5 = 1/243

Hence, 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:

an = a+(n-1)d

a10 = 14 + (10 – 1)(-4)

a10 = 14 + (9)(-4)

a10 = 14 – 36 = -22

Hence, the 10th term of the sequence is -22.

Problem 3: If p, q, and r are in A.P., find the value of (q2-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 = 4d2

q2 = a2

p × r = (a – d) (a + d) = (a2 – d2)

q2 – pr = a2 – (a2 – d2) = d2

So, (q2 – pr)/(p – q)2 = d2/4d2 = 1/4

Hence, the value (q2-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) = 2 and common difference (d) = 2

Now,

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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