How much of JEE Main comes from Sequences and Series?

JEEGenie estimates about 4% of JEE Main marks come from Sequences and Series, based on recent papers. The share varies from year to year.

Which formulas matter most in Sequences and Series?

The Sequences and Series sheet on this page has 19 formula entries, starting with Arithmetic Progression (AP), Sum of n Terms of an AP, Geometric Progression (GP), Sum of n Terms of a GP, Sum of Infinite GP. Each one comes with when to use it and the trap to avoid.

What topics does Sequences and Series cover for JEE?

Sequences and Series covers Arithmetic Progression, Geometric Progression, Harmonic Progression, AM-GM-HM Inequality, Sum of Special Series, Arithmetic-Geometric Series.

How should I revise Sequences and Series before the exam?

Learn each formula together with its condition of use, read the trap listed beside it, then practise questions and review every mistake. The practice link on this page opens Sequences and Series questions on JEEGenie.

How many previous-year questions are there on Sequences and Series?

JEEGenie has 538 previous-year questions on Sequences and Series, and this page shows a few with worked explanations.

Is practice on Sequences and Series free on JEEGenie?

Sign-up is free and includes a 7-day Pro trial. After the trial, free accounts keep a limited daily quota.

Sequences and Series for JEE Main & Advanced — formulas, what is asked, and traps

JEEGenie estimates about 4% of JEE Main marks come from Sequences and Series. This page covers 19 formula entries, each with when to use it, what JEE asks and the traps. A common slip: Confusing sequence (list) with series (sum) is a common error.

Key points to know

Sequence and Series. A sequence is an ordered list of numbers following a rule; a series is the sum of terms of a sequence.
Arithmetic Progression (AP). A sequence where each term differs from the previous by a constant common difference d.
Geometric Progression (GP). A sequence where each term is obtained by multiplying the previous by a constant common ratio r.
Arithmetic Mean (AM). The arithmetic mean of two numbers a and b is their average.
Geometric Mean (GM). The geometric mean of two positive numbers a and b is the square root of their product.
Harmonic Mean (HM). The harmonic mean of two positive numbers a and b is the reciprocal of the arithmetic mean of their reciprocals.
Harmonic Progression (HP). A sequence is in HP if the reciprocals of its terms are in AP.

Formula sheet

Name	Formula	When to use
Arithmetic Progression (AP)	$a_{n} = a + (n - 1) d$	When terms increase or decrease by a fixed amount.
Sum of n Terms of an AP	$S_{n} = \frac{n}{2} [2 a + (n - 1) d] = \frac{n}{2} (a + l)$ where $l$ is last term	To find sum of first n terms or to find n when sum is given.
Geometric Progression (GP)	$a_{n} = a r^{n - 1}$	When terms multiply by a fixed factor.
Sum of n Terms of a GP	$S_{n} = \frac{a (1 - r^{n})}{1 - r}$ for $r \neq 1$ ; $S_{n} = n a$ for $r = 1$	To sum a finite number of terms in a GP.
Sum of Infinite GP	$S_{\infty} = \frac{a}{1 - r}$ for $∣ r ∣ < 1$	When a GP has infinite terms and \|r\| < 1, e.g., recurring decimals.
Arithmetic Mean (AM)	$A M = \frac{a + b}{2}$	To find the average or insert arithmetic means between numbers.
Geometric Mean (GM)	$G M = \sqrt{a b}$	To find the geometric mean or insert geometric means between numbers.
Harmonic Mean (HM)	$H M = \frac{2 a b}{a + b}$	To find the harmonic mean or insert harmonic means, often in speed or rate problems.
Relation between AM, GM, HM	$A M \geq G M \geq H M$	To compare means or prove inequalities in sequences.
Sum of Natural Numbers	$\sum_{k = 1}^{n} k = \frac{n (n + 1)}{2}$	When summing consecutive natural numbers starting from 1.
Sum of Squares of Natural Numbers	$\sum_{k = 1}^{n} k^{2} = \frac{n (n + 1) (2 n + 1)}{6}$	When summing squares of consecutive natural numbers.
Sum of Cubes of Natural Numbers	$\sum_{k = 1}^{n} k^{3} = {[\frac{n (n + 1)}{2}]}^{2}$	When summing cubes of consecutive natural numbers.
Summation Using Partial Fractions	$\frac{1}{k (k + 1)} = \frac{1}{k} - \frac{1}{k + 1}$	When the general term is a rational function that can be decomposed, e.g., $\frac{1}{n (n + 1)}$ .
Harmonic Progression (HP)	$\frac{1}{a_{n}}$ forms an AP	When terms are reciprocals of an AP, e.g., $1, 1 / 2, 1 / 3, . . .$ .
Arithmetico-Geometric Progression (AGP)	$S_{n} = \frac{a}{1 - r} + \frac{d r (1 - r^{n - 1})}{(1 - r)^{2}} - \frac{(a + (n - 1) d) r^{n}}{1 - r}$ for $r \neq 1$	When each term is of the form (linear in n) * (geometric term), e.g., $1 \cdot 2 + 2 \cdot 2^{2} + 3 \cdot 2^{3} + . . .$ .
Sum of Special Series	$\sum (a n^{2} + b n + c) = a \sum n^{2} + b \sum n + c \sum 1$	When the general term is a polynomial in n, e.g., $\sum (3 n^{2} + 2 n + 1)$ .
Inserting Means Between Numbers	$d = \frac{b - a}{n + 1}$ , $r = {(\frac{b}{a})}^{\frac{1}{n + 1}}$	When asked to insert a given number of means between two numbers.
Sigma Notation and Properties	$\sum_{k = 1}^{n} (a_{k} \pm b_{k}) = \sum a_{k} \pm \sum b_{k}$	To manipulate and simplify series sums efficiently.
Convergence of Infinite Series	$\sum_{n = 1}^{\infty} a r^{n - 1}$ converges iff $∣ r ∣ < 1$	To determine if an infinite sum exists, especially in geometric series.

Open the Sequences and Series formula page

What JEE asks

Sequence and Series. JEE often tests the distinction in word problems: 'find the sum' vs 'find the nth term'.
Arithmetic Progression (AP). JEE frequently asks for the nth term or sum given two terms, or to find d from conditions.
Sum of n Terms of an AP. JEE often uses $S_{n}$ to derive relationships between terms, e.g., $S_{n} - S_{n - 1} = a_{n}$ .
Finding AP from Conditions. JEE frequently uses symmetric assumption to reduce equations, e.g., sum of three terms = 15, product = 80.
Geometric Progression (GP). JEE tests nth term and sum, especially with r between -1 and 1 for infinite sums.
Sum of n Terms of a GP. JEE often combines GP sum with other series or uses it in word problems like population growth.
Sum of Infinite GP. JEE asks for sum of infinite GP in geometric series problems, often with decimals or geometric constructions.
Finding GP from Conditions. JEE often uses this to solve for a and r quickly, e.g., product of three terms = 27, sum = 13.
Arithmetic Mean (AM). JEE tests AM in relation to GM and HM, often in inequality problems.
Geometric Mean (GM). JEE uses GM in problems involving growth rates or ratios.
Harmonic Mean (HM). JEE tests HM in relation to AM and GM, especially in inequalities.
Relation between AM, GM, HM. JEE frequently uses this in multiple-choice questions to determine the largest or smallest mean.
Sum of Natural Numbers. JEE uses this as a building block for more complex series sums.
Sum of Squares of Natural Numbers. JEE often combines this with other sums in series problems.
Sum of Cubes of Natural Numbers. JEE uses this to simplify series where terms are cubic expressions.
Method of Differences. JEE often tests this for finding nth term or sum of series like $1 \cdot 2 + 2 \cdot 3 + 3 \cdot 4 + . . .$ .
Summation Using Partial Fractions. JEE frequently uses this for sums like $\sum \frac{1}{n (n + 1)}$ or $\sum \frac{1}{n (n + 2)}$ .
Harmonic Progression (HP). JEE tests HP by converting to AP, e.g., find nth term of $1, 1 / 3, 1 / 5, . . .$ .
Arithmetico-Geometric Progression (AGP). JEE often asks for sum to n terms or infinite sum (if |r|<1) of AGP.
Sum of Special Series. JEE tests this by giving series like $\sum (n^{3} + 2 n)$ and asking for sum.
Inserting Means Between Numbers. JEE often asks to find the means or the common difference/ratio.
Sigma Notation and Properties. JEE tests this in problems where series are expressed in sigma form and need to be evaluated.
Convergence of Infinite Series. JEE tests convergence in context of infinite GP sums and sometimes in more complex series.

Common traps

Confusing sequence (list) with series (sum) is a common error. Sequence and Series
Using $a_{n} = a + n d$ instead of $a + (n - 1) d$ . Arithmetic Progression (AP)
Using $S_{n} = n / 2 * (a + l)$ when l is not the last term of n terms. Sum of n Terms of an AP
Using $a, a + d, a + 2 d$ for three terms when sum is given – leads to more algebra. Finding AP from Conditions
Using $a_{n} = a r^{n}$ instead of $a r^{n - 1}$ . Geometric Progression (GP)
Using the formula when r=1 leads to division by zero. Sum of n Terms of a GP
Applying formula when |r| >= 1 – series diverges, no finite sum. Sum of Infinite GP
Using $a, a r, a r^{2}$ when product is given – leads to more algebra. Finding GP from Conditions
Confusing AM with the middle term of an AP – AM of $n$ numbers is their average; between $a$ and $b$ the single AM is the middle term of $a$ , AM, $b$ . Arithmetic Mean (AM)
Applying GM to negative numbers – defined only for non-negative. Geometric Mean (GM)
Forgetting that HM is defined only for non-zero numbers. Harmonic Mean (HM)
Applying the inequality to non-positive numbers. Relation between AM, GM, HM
Forgetting the factor 1/2 – common mistake. Sum of Natural Numbers
Forgetting the factor 1/6 – common error. Sum of Squares of Natural Numbers

Topics in Sequences and Series

Arithmetic Progression
Geometric Progression
Harmonic Progression
AM-GM-HM Inequality
Sum of Special Series
Arithmetic-Geometric Series

Previous-year questions on Sequences and Series

JEEGenie has 538 previous-year questions on Sequences and Series. A few worked examples follow.

Year	Questions
2026	34
2025	28
2024	38
2023	40
2022	39
2021	45
2020	29
2019	62
2018	16
2017	11
2016	13
2015	10
2014	9
2013	13
2012	10
2011	4
2010	4
2009	2
2008	1
2007	11
2006	6
2005	5
2004	7
2003	3
2002	11
2001	6
2000	7
1999	9
1998	5
1997	6
1996	2
1995	3
1994	3
1993	1
1992	1
1991	7
1990	1
1989	1
1988	5
1987	2
1986	4
1985	4
1984	4
1983	1
1982	4
1981	1
1980	2
1979	4
1978	4

Question 1 medium

Let A be the set of first 101 terms of an A.P., whose first term is 1 and the common difference is 5 and let B be the set of first 71 terms of an A.P., whose first term is 9 and the common difference is 7. Then the number of elements in

A \cap B

, which are divisible by 3, is :

Answer: B

Explanation. A = {1, 6, 11,

\dots

, 1 + 5(101−1)} = {1 + 5(k−1) : 1

\leq

\leq

101} B = {9, 16, 23,

\dots

, 9 + 7(71−1)} = {9 + 7(m−1) : 1

\leq

\leq

71} Common terms satisfy

1 + 5 (k - 1) = 9 + 7 (m - 1) \Rightarrow 5 k - 7 m = 6

. Solve:

5 k \equiv 6 (m o d 7) \Rightarrow 5 k \equiv 6 (m o d 7) \Rightarrow k \equiv 4 (m o d 7)

(since

5^{- 1} \equiv 3

). So

k = 4 + 7 t

. Then

5 (4 + 7 t) - 7 m = 6 \Rightarrow 20 + 35 t - 7 m = 6 \Rightarrow 7 m = 14 + 35 t \Rightarrow m = 2 + 5 t

. Constraints:

1 \leq k \leq 101 \Rightarrow 1 \leq 4 + 7 t \leq 101 \Rightarrow 0 \leq t \leq 13

1 \leq m \leq 71 \Rightarrow 1 \leq 2 + 5 t \leq 71 \Rightarrow 0 \leq t \leq 13

. Thus

t = 0, 1, \dots, 13

gives 14 common terms. Now require common term divisible by 3. Common term

= 1 + 5 (k - 1) = 1 + 5 (3 + 7 t) = 16 + 35 t

. Divisible by 3:

16 + 35 t \equiv 0 (m o d 3) \Rightarrow 1 + 2 t \equiv 0 (m o d 3) \Rightarrow 2 t \equiv 2 (m o d 3) \Rightarrow t \equiv 1 (m o d 3)

. So

t = 1, 4, 7, 10, 13

— that’s 5 values. **Answer: B) 5**

From NTA/IIT papers: JEE Main 2026 · 2 Apr Shift 1

Question 2 hard

Let

R

denote the set of all real numbers. Let

f : R \to R

be a function such that

f (x) > 0

for all

x \in R

, and

f (x + y) = f (x) f (y)

for all

x, y \in R

. Let the real numbers

a_{1}, a_{2}, \dots, a_{50}

be in an arithmetic progression. If

f (a_{31}) = 64 f (a_{25})

, and

\sum_{i = 1}^{50} f (a_{i}) = 3 (2^{25} + 1),

then the value of

\sum_{i = 6}^{30} f (a_{i})

is _______.

Answer: 96

Explanation. Since

f (x + y) = f (x) f (y)

and

f (x) > 0

f

is an exponential function:

f (x) = b^{x} (b > 0) .

Let the arithmetic progression be

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

Then

f (a_{n}) = b^{A + (n - 1) d} = b^{A} {(b^{d})}^{n - 1} .

Let

r = b^{d} > 0, C = b^{A} > 0.

f (a_{n}) = C r^{n - 1} .

Given

f (a_{31}) = 64 f (a_{25}),

we have

C r^{30} = 64 C r^{24} .

Since

C > 0

r^{6} = 64 = 2^{6},

r = 2.

Thus

f (a_{n}) = C \cdot 2^{n - 1} .

Now

\sum_{i = 1}^{50} f (a_{i}) = C \sum_{i = 1}^{50} 2^{i - 1} = C (2^{50} - 1) .

Given

C (2^{50} - 1) = 3 (2^{25} + 1) .

Factor:

2^{50} - 1 = (2^{25} - 1) (2^{25} + 1) .

C (2^{25} - 1) (2^{25} + 1) = 3 (2^{25} + 1) .

Since

2^{25} + 1 \neq 0

C (2^{25} - 1) = 3.

Hence

C = \frac{3}{2^{25} - 1} .

We need

\sum_{i = 6}^{30} f (a_{i}) = C \sum_{i = 6}^{30} 2^{i - 1} .

The sum is

\sum_{i = 6}^{30} 2^{i - 1} = 2^{5} + 2^{6} + \dots + 2^{29} .

This is a geometric series with first term

2^{5}

, ratio

2

, and

25

terms:

2^{5} \frac{2^{25} - 1}{2 - 1} = 2^{5} (2^{25} - 1) .

Therefore

\sum_{i = 6}^{30} f (a_{i}) = C \cdot 2^{5} (2^{25} - 1) = \frac{3}{2^{25} - 1} \cdot 32 (2^{25} - 1) = 96.

96.0

From NTA/IIT papers: JEE Advanced 2025

Question 3 easy

The 20^{th} term from the end of the progression 20, 19 \frac{1}{4}, 18 \frac{1}{2}, 17 \frac{3}{4}, \dots, - 129 \frac{1}{4} is :

$- 115$
$- 100$
$- 110$
$- 118$

Answer: A

Explanation. The sequence is arithmetic with first term

a = 20

and common difference

d = 19 \frac{1}{4} - 20 = - \frac{3}{4}

. Last term

l = - 129 \frac{1}{4} = - \frac{517}{4}

. The

20^{th}

term from the end is the

(n - 19)^{th}

term from the start. First find

n

l = a + (n - 1) d ⟹ - \frac{517}{4} = 20 + (n - 1) (- \frac{3}{4})

- \frac{517}{4} = \frac{80}{4} - \frac{3}{4} (n - 1) ⟹ - 517 = 80 - 3 (n - 1)

- 597 = - 3 (n - 1) ⟹ n - 1 = 199 ⟹ n = 200

The

20^{th}

term from the end is the

(200 - 20 + 1) = 181^{st}

term.

T_{181} = 20 + (181 - 1) (- \frac{3}{4}) = 20 - 180 \cdot \frac{3}{4} = 20 - 135 = - 115

The answer is option A.

From NTA/IIT papers: JEE Main 2024 · 27 Jan Shift 2

Question 4 medium

The sum of the first

20

terms of the series

5 + 11 + 19 + 29 + 41 + \dots

is :

3420
3450
3250
3520

Answer: D

Explanation. Find the pattern of the series: 5, 11, 19, 29, 41,... Differences between consecutive terms: 6, 8, 10, 12,... (arithmetic progression with common difference 2) Let

T_{n}

be the

n

-th term. Using the method of differences:

T_{n} = T_{1} + \sum_{k = 1}^{n - 1} (2 k + 4) = 5 + \sum_{k = 1}^{n - 1} (2 k + 4) = 5 + 2 \cdot \frac{(n - 1) n}{2} + 4 (n - 1) = 5 + n (n - 1) + 4 (n - 1) = 5 + (n - 1) (n + 4) = 5 + n^{2} + 3 n - 4 = n^{2} + 3 n + 1

Verification:

T_{1} = 1 + 3 + 1 = 5

✓,

T_{2} = 4 + 6 + 1 = 11

✓,

T_{3} = 9 + 9 + 1 = 19

✓ Sum of first 20 terms:

S_{20} = \sum_{n = 1}^{20} (n^{2} + 3 n + 1) = \sum_{n = 1}^{20} n^{2} + 3 \sum_{n = 1}^{20} n + \sum_{n = 1}^{20} 1 = \frac{20 \cdot 21 \cdot 41}{6} + 3 \cdot \frac{20 \cdot 21}{2} + 20 = \frac{17220}{6} + 3 \cdot 210 + 20 = 2870 + 630 + 20 = 3520

From NTA/IIT papers: JEE Main 2023 · 6 Apr Shift 1

Question 5 hard

Let

l_{1}, l_{2}, \dots, l_{100}

be consecutive terms of an arithmetic progression with common difference

d_{1}

, and let

w_{1}, w_{2}, \dots, w_{100}

be consecutive terms of another arithmetic progression with common difference

d_{2}

, where

d_{1} d_{2} = 10

. For each

i = 1, 2, \dots, 100

, let

R_{i}

be a rectangle with length

l_{i}

, width

w_{i}

and area

A_{i}

. If

A_{51} - A_{50} = 1000

, then the value of

A_{100} - A_{90}

is ______.

Answer: 18900

Explanation. We have

l_{i} = l_{1} + (i - 1) d_{1}, w_{i} = w_{1} + (i - 1) d_{2} .

Thus

A_{i} = l_{i} w_{i} .

Compute

A_{i + 1} - A_{i} = (l_{i} + d_{1}) (w_{i} + d_{2}) - l_{i} w_{i} = l_{i} d_{2} + w_{i} d_{1} + d_{1} d_{2} .

Given

d_{1} d_{2} = 10

and

A_{51} - A_{50} = 1000

, we have

l_{50} d_{2} + w_{50} d_{1} + 10 = 1000 \Rightarrow l_{50} d_{2} + w_{50} d_{1} = 990.

Now

A_{100} - A_{90} = \sum_{i = 90}^{99} (A_{i + 1} - A_{i}) .

For each

i

A_{i + 1} - A_{i} = l_{i} d_{2} + w_{i} d_{1} + 10.

Since

l_{i} = l_{50} + (i - 50) d_{1}

and

w_{i} = w_{50} + (i - 50) d_{2}

l_{i} d_{2} + w_{i} d_{1} = (l_{50} d_{2} + w_{50} d_{1}) + (i - 50) (d_{1} d_{2} + d_{2} d_{1}) = 990 + (i - 50) (20) .

A_{i + 1} - A_{i} = 990 + 20 (i - 50) + 10 = 1000 + 20 (i - 50) .

Summing from

i = 90

99

A_{100} - A_{90} = \sum_{i = 90}^{99} [1000 + 20 (i - 50)] .

There are 10 terms.

\sum_{i = 90}^{99} 1000 = 10000,

and

\sum_{i = 90}^{99} 20 (i - 50) = 20 \sum_{i = 90}^{99} (i - 50) = 20 \sum_{k = 40}^{49} k .

\sum_{k = 40}^{49} k = \frac{(40 + 49) \cdot 10}{2} = 445.

Thus

20 \cdot 445 = 8900.

Therefore

A_{100} - A_{90} = 10000 + 8900 = 18900.

18900

From NTA/IIT papers: JEE Advanced 2022

Frequently asked questions

How much of JEE Main comes from Sequences and Series?: JEEGenie estimates about 4% of JEE Main marks come from Sequences and Series, based on recent papers. The share varies from year to year.
Which formulas matter most in Sequences and Series?: The Sequences and Series sheet on this page has 19 formula entries, starting with Arithmetic Progression (AP), Sum of n Terms of an AP, Geometric Progression (GP), Sum of n Terms of a GP, Sum of Infinite GP. Each one comes with when to use it and the trap to avoid.
What topics does Sequences and Series cover for JEE?: Sequences and Series covers Arithmetic Progression, Geometric Progression, Harmonic Progression, AM-GM-HM Inequality, Sum of Special Series, Arithmetic-Geometric Series.
How should I revise Sequences and Series before the exam?: Learn each formula together with its condition of use, read the trap listed beside it, then practise questions and review every mistake. The practice link on this page opens Sequences and Series questions on JEEGenie.
How many previous-year questions are there on Sequences and Series?: JEEGenie has 538 previous-year questions on Sequences and Series, and this page shows a few with worked explanations.
Is practice on Sequences and Series free on JEEGenie?: Sign-up is free and includes a 7-day Pro trial. After the trial, free accounts keep a limited daily quota.