Sequence and Series

A sequence is defined as the succession of numbers formed according to some definite rules. A clear introduction to sequence and series is given in this article.

A sequence is a group of numbers in an ordered form that obeys a certain pattern. The sequence and series can be finite or infinite depending upon the number of terms in finite or infinite, respectively. A series can be constructed by adding or subtracting the terms of a sequence.  The n^th term is denoted by a_n or T_n, and it is the general term of sequence/series.

A finite sequence is a sequence that has a finite number of terms. For example, {2, 4, 6, 8} is a sequence of multiples of 2 less than 9. 

An infinite sequence has an unlimited number of terms. For example, 3, 6, 9,… 

Types of Sequence and Series

There are three types of sequence, as mentioned below:

• Arithmetic Sequence
• Geometric Sequence
• Fibonacci Sequence

And mainly, we deal with two types of series:

• Arithmetic Series
• Geometric Series

Arithmetic Progression

A sequence of numbers is called an arithmetic progression when there is a common difference between two consecutive terms.

Important Formulas

n^th term of AP = a + (n – 1) d

(n+1)^th term = a + nd

Where, a is the first term and d is the common difference.

Properties of Arithmetic sequences

1) The resulting sequence of an AP remains the same if a constant is added or subtracted from each term of an AP.

2) If we select terms in the regular interval from an A.P., these selected terms will also be in AP.

3) The resulting sequence of an AP remains the same if a constant non-zero number is divided or multiplied by each term of an AP.

4) Selection of terms in AP

3 terms: a – d, a, a + d

4 terms: a – 3d, a – d, a + d, a + 3d

5 terms: a – 2d, a – d, a, a + d, a + 2d

Geometric Progression

A sequence of numbers is a geometric progression when there is a common ratio that comes by multiplying or dividing the two consecutive terms.

General Form: a, ar, ar², ar³, …, arⁿ

n^th term of a GP = ar^n-1

Sum of n terms of a GP = a(1 – rⁿ)/(1 – r) {when r < 1 and r ≠ 1}

(or)

S_n = a(rⁿ – 1)/(r – 1) {when r > 1 and r ≠ 1}

Where, r is the common ratio, and a is the first term.

Fibonacci Sequence

A sequence of numbers is called a Fibonacci sequence when the required term is obtained by adding the two terms before that.

What Is a Series in Maths?

The summation of all the terms of the sequence is called a series. It is denoted as S_n. For example, If we are given the sequence 2, 4, 6, 8, …., Then the series is 2 + 4 + 6 + 8 + ….

Important summations

The sum of first n natural numbers:

The sum of squares of first n natural numbers:

The sum of cubes of first n natural numbers:

Also,

Solved Problems for IIT JEE

Problem 1:

Find the nth term of the given series 1, (1 + 2), (1 + 2 + 3), (1 + 2 + 3 + 4), . . . . . . . . . . . . . .

Solution:

Here, n^th term = (1 + 2 + 3 + 4 + 5 . . . . . . . . . . . .+ n)

This is an arithmetic series with first term = 1 and common difference = 1

Then,

Which is the n^th term of the given series.

Problem 2:

If the 9th term of an AP is zero, find the ratio of its 29th term and 19th term.

Solution:

9th term = a + (9 – 1) d = 0

⇒ a + 8d = 0

or a = -8d

Now,

(29th term)/(19th term) = (a + 28d) / (a + 18d) = ((-8d + 28d)) / ((-8d + 18d)) = 20d / 10d = 2/1

Thus, the ratio is 2:1.

Problem 3:

If the nth term of an AP is (2n – 1), then find the sum of its n terms.

Solution: We know,

= 2 (n (n + 1) / 2) – n

= n²

Problem 4:

If the nth term of the series 63, 65, 67, 69 . . . . . . . . . . . . and mth term of the series 3, 10, 17, 24 . . . . . . . . . . . . . . . . are equal, then find the relation between m and n.

Solution: The nth term of the first series = 63 + (n – 1)2 = 61 + 2n

The mth term of the second series = 3 + (m – 1)7 = 7m – 4

If both are equal, 61 + 2n = 7m – 4

or 7m – 2n = 65

Problem 5:

If a, b, and c are in AP, then find the value of (a-c)² / (b²-ac).

Solution:

Given, a, b, and c are in AP.

So, 2b = a + c

b = (a + c)/2

Now,

b² – ac = [(a + c)/2]² – ac

= (a² + c² + 2ac – 4ac)/4

= (a – c)²/4

Therefore, (a-c)² / (b²-ac) = 4(a – c)² / (a – c)²

= 4

Problem 6:

Find the sum of the first 20 terms of the sequence 0.7, 0.77, 0.777 . . . . . . . . up to 20 terms.

Solution:

0.7 + 0.77 + 0.777 + …. up to 20 terms

= 7(0.1 + 0.11 + 0.111 + …. up to 20 terms)

Problem 7:

Let a₁, a₂, a₃,… be a harmonic progression with its first term as 5 and 20th term as 25. The least positive integer n for which a_n < 0 is _____.

Solution:

If a₁, a₂, a₃,… be a harmonic progression [HP], then 1/a₁, 1/a₂, 1/a₃,… will be in arithmetic progression [AP].

First-term of AP,

Term 20 of AP,

We have to find the least positive integer n for which 

So, n = 25

Problem 8:

Find the sum of all natural numbers lying between 101 and 199, which are multiples of 5.

Solution:

We know the nearest number to 101, which is a multiple of 5 and lies between 101 and 199, is 105. Similarly, 195 is the nearest number to 199 which is a multiple of 5, which lies between 101 and 199. So, we take the last term of AP as 195.

We have the first term, a = 105 

The common difference, d = 5

Last term, l = 195

l = a + (n – 1) d

195 = 105 + (n – 1)5

195 – 105 = ( n – 1)5

5n – 5 = 90

5n = 90+5 = 95

n = 95/5 = 19

Sum of n terms = (n/2)(a + l)

= (19/2)(105 + 195) = 2850

Hence, the required sum is 2850.

Sequence and Series – Top 12 Important and Expected JEE Questions

Also Check:

Arithmetic Progression Solved Examples

Geometric Progression Solved Examples

Top Questions on Sequence and Series for JEE Advanced

Sequence and Series – Rapid Revision 1

Sequence and Series – Rapid Revision 2

Sequence and Series – Rapid Revision 3

Sequence and Series Class 11 – Bridge Course

Sequence and Series – JEE Important Questions

 Telescopic Series