Sequence and Series Formulas for JEE Main 2024 - Download PDF

The sequence and series formulas for the JEE Main exam are very few. These formulas must be remembered to be able to answer all the questions on sequence and series in the JEE Main Mathematics syllabus. Sequence and series carry a weightage of 6.6% in the JEE Main 2024 exam.

Table of Contents

• Sequence and Series Formulas for JEE Main 2024 - Download PDF
• Sequence and Series Formulas for JEE Main 2024- Topic-wise
• JEE Main 2024 Sequence and Series Questions for Practice
• JEE Main Sequence and Series Weightage

Sequence and Series Formulas for JEE Main 2024 - Download PDF

The Sequence and Series is a part of the Mathematics Syllabus of the JEE Main examination. JEE Main Mathematics syllabus includes Arithmetic Sequence, Geometric Sequence, Harmonic Sequence, and Fibonacci Number under the ‘Sequence and Series’ topic.

Shared below is a PDF containing the important formulas for Sequence and Series:

Also Read: JEE Main Important Formulas 2024

Sequence and Series Formulas for JEE Main 2024 - Topic-wise

Sequence and Series carry moderate weightage in the JEE Main exams. It is equally important for students to check and understand the JEE Main syllabus as a whole along with specific subject-wise. Students can check the formulas given below:

1. Arithmetic Sequence
Arithmetic Sequence Formula for nth Term: a_n = a + (n-1)d
Arithmetic Series Formula for nth Term: s_n = n/2 {2a + (n-1)d}

2. Geometric Sequence
Geometric Sequence Formula for nth Term: a_n = ar^(n-1)
Geometric Series Formula for nth Term: s_n = {a(1-rⁿ)} / (1-r)
Geometric Series Formula for Infinite Term: s_n = a / (1-r), where |r| < 1

Also Read: Most Easy and Scoring Chapters in JEE Main 2024

3. Harmonic Sequences
Harmonic Sequence Formula for nth Term: a_n = 1 / {a + (n-1)d}
Harmonic Series Formula for nth Term: s_n = 1/d In [ {2a + (2n-1)d} / (2a - d)]

4. Fibonacci Number Sequence
The formula for the Fibonacci Sequence is
F₁= 0,
F₂ = 1,
F₃ = 1,
F₄ = 2,
...
F_n= F_n-1 + F_n-2

Also Read: JEE Main Maths Important Formulas 2024 - Download PDF

JEE Main 2024 Sequence and Series - Questions for Practice

Given below are a few important questions from the topic - sequence and series - that are asked in the JEE Main exam. The students can practice these questions and other questions similar to these.

Question 1: The interior angles in a polygon are in arithmetic progression. If the smallest angle is 120° and the common difference is 50°, what is the number of sides of the polygon?

Solution: Let the number of sides of the given polygon be x.

The sum of the interior angles of the polygon = (2x − 4) [π / 2] = (x − 2)π

Since the angles are in A.P. and a = 120° , d = 5, therefore

[x / 2] × [2 × 120 + (x − 1)5] = (x − 2) × 180

x2− 25x + 144 = 0

(x − 9) (x − 16) = 0

x = 9, 16

When x = 16,

T16 = a + 15d = 120° + 15.5° = 195°, which is not possible because an interior angle cannot be more than 180°.

Hence, x = 9.

Also Read: JEE Main Chapter Wise Weightage

Question 2. x + y + z = 15; if 9, x, y, z, a are in A.P.; while [1 / x] + [1 / y] + [1 / z] = 5 / 3 if 9, x, y, z, a are in H.P., then what is the value of a?

Solution: x + y + z = 15, if x = (z-3)-1 = z3 are in A.P.

Sum =9+15+a=52(9+a)

⇒ 24 + a = 5 / 2 (9 + a)

⇒ 48 + 2a = 45 + 5a

⇒ 3a = 3

⇒ a = 1 ..(i) and

[1 / x] + [1 / y] + [1 / z] = 5 / 3, if 9, x, y, z, are in H.P.

Sum = 1 / 9 + 5 / 3 + 1 / a

= 5 / 2 [1 / 9 + 1 / a]

⇒ a = 1 

Also Read: Important Topics for JEE Main Session 2 2024

Question 3. The sum of n terms of the following series 1 + (1 + x) + (1 + x + x2) + . . . . . will be ___________.

Solution: 1 + (1 + x) + (1 + x + x2) + . . . . .+ (1 + x + x2+ x3+ . . . + xn-1) + . . .

Required sum = [1 / (1 − x)] * {(1 − x) + (1 − x2) + (1 − x3) + (1 − x4) + ……….upto n terms}

= [1 / (1 − x)] * [n − {x + x2 + x3 + . . . . . . . . . . till n terms } ]

= [1 / (1 − x)] * [n − {x (1 − xn) / [1 − x]}]

= [n (1 − x) − x (1 − xn)] / [(1-x)2]

Also Read: Most Easy Chapters of JEE Main 2024

Question 4. The sum of the series 1 + [1] / [4 × 2!] + [1] / [16 × 4!] + [1] / [64 × 6!] + . . . . . is ________.

Solution: [ex + e-x] / [2] = 1 + [x2/ 2!] + [x4/ 4!] + [x6 / 6!] + . . . . ∞

Putting x = 1 / 2, we get 1 + [1] / [4 × 2!] + [1] / [16 × 4!] + [1] / [64 × 6!] + . . . . . ∞

= [e1/2 + e-1/2 ] / [2]

= [e + 1] / 2√e

 Also Read: Revision Notes for JEE Main 2024

JEE Main Sequence and Series Weightage

Sequence and Series hold a moderate weightage in the JEE Main exam. However, the students must practice the sums on sequence and series to ensure full marks on this topic. The weightage of Sequence and Series has been shared below:

Also Read:JEE Main Chapter Wise Weightage 2024