Sequence and Series Formula Sheet — JEE Main Mathematics

Every key Sequence and Series formula, definition and theorem for JEE Main Mathematics in one place — with common examiner traps and worked examples. Free to read; blurt from memory, then check your gaps.

Syllabus — topics coveredNTA · 3 sub-topics

Arithmetic and Geometric progressions
Insertion of arithmetic and geometric means between two given numbers
Relation between A.M and G.M

Sequences & Arithmetic Progression (A.P.)

Arithmetic progression: A sequence

a, a + d, a + 2 d, \dots

with a

constant common difference

d. The n-th term is

a_{n} = a + (n - 1) d

and the sum of n terms is

S_{n} = \frac{n}{2} [2 a + (n - 1) d] = \frac{n}{2} (a + l)

, where l is the last term.

Working with an A.P.

▸ $d = a_{n} - a_{n - 1}$ is constant; $a_{n} = S_{n} - S_{n - 1}$ recovers a term from the sums.
▸Three terms in A.P.: take $a - d, a, a + d$ (sum $= 3 a$ ); five terms: $a - 2 d, \dots, a + 2 d$ .
▸Terms equidistant from the ends sum to a constant: $a_{k} + a_{n - k + 1} = a + l$ .
▸ $a_{n}$ is the average of its neighbours: $a_{n} = \frac{a _{n - 1} + a _{n + 1}}{2}$ .

🎯 Exam · Choose symmetric terms

When the

sum

of terms in A.P. is given, pick them symmetrically about the middle (

a - d, a, a + d

) so the d's cancel and the sum gives a directly. This is the single most useful A.P. trick.

★ Remember · Sₙ is quadratic in n

S_{n} = \frac{d}{2} n^{2} + (a - \frac{d}{2}) n

is a quadratic in n with no constant term. Conversely, if

S_{n} = A n^{2} + B n

then the sequence is an A.P. with

d = 2 A

⚠️ Watch out · (n−1)d, not nd

The

n

-th term is

a + (n - 1) d

— the first term has

n = 1

and no

d

. Using

a + n d

shifts everything by one term, a very common slip.

🚫 Examiner Trap · Examiner traps

(1) n-th term uses

(n - 1) d

, not nd:

a_{n} = a + (n - 1) d

. (2) For 3 terms in A.P. take

a - d, a, a + d

(sum

3 a

) — symmetric choice simplifies algebra; for 4 take

a - 3 d, a - d, a + d, a + 3 d

(common diff

2 d

). (3) Terms equidistant from the ends have a CONSTANT sum

a_{k} + a_{n - k + 1} = a + l

. (4)

S_{n} = \frac{n}{2} (a + l)

needs the LAST term l; or

\frac{n}{2} [2 a + (n - 1) d]

Geometric Progression (G.P.)

Geometric progression: A sequence

a, a r, a r^{2}, \dots

with a

constant common ratio

r. The n-th term is

a_{n} = a r^{n - 1}

and the sum is

S_{n} = \frac{a ( r ^{n} - 1 )}{r - 1}

for

r \neq = 1

(and

S_{n} = na

r = 1

★ Remember · Infinite G.P.

S_{\infty} = \frac{a}{1 - r}

, valid

only

for

∣ r ∣ < 1

(otherwise the terms don't shrink and the sum diverges). This converts a repeating decimal to a fraction:

0. \overline{3} = \frac{3/10}{1 - 1/10} = \frac{1}{3}

G.P. properties

▸Three terms in G.P.: take $\frac{a}{r}, a, a r$ (product $= a^{3}$ ).
▸Each term squared $=$ product of its neighbours: $a_{n}^{2} = a_{n - 1} a_{n + 1}$ .
▸Terms equidistant from the ends have a constant product.
▸Logs of a G.P. (with positive terms) form an A.P.

🎯 Exam · Geometric mean is the bridge

Any term of a G.P. is the geometric mean of the terms either side:

b = a c

for consecutive a,b,c. Three numbers are in G.P.

\Leftrightarrow b^{2} = a c

⚠️ Watch out · Check |r| < 1 before using S∞

The infinite-sum formula only applies when

∣ r ∣ < 1

. If

∣ r ∣ \geq 1

there is no finite sum — a frequent trap when r involves an unknown that could be large.

🚫 Examiner Trap · Examiner traps

(1)

S_{n} = \frac{a ( r ^{n} - 1 )}{r - 1}

holds only for

r \neq = 1

; if

r = 1

then

S_{n} = na

. (2) Infinite sum

\frac{a}{1 - r}

exists ONLY when

∣ r ∣ < 1

(else it diverges). (3) For 3 terms in G.P. take

\frac{a}{r}, a, a r

(product

a^{3}

). (4) Each term

^{2} =

product of its neighbours:

a_{n}^{2} = a_{n - 1} a_{n + 1}

— but watch the SIGN of r when taking square roots.

Harmonic Progression & The Three Means

Harmonic progression: A sequence is in

H.P.

exactly when the reciprocals of its terms form an

A.P.

There is no neat sum formula, so the standard move is to flip to the reciprocal A.P. and solve there.

The three means of a, b > 0

▸Arithmetic mean $A = \frac{a + b}{2}$ .
▸Geometric mean $G = ab$ .
▸Harmonic mean $H = \frac{2 ab}{a + b}$ .
▸H is the reciprocal of the A.M. of $\frac{1}{a}, \frac{1}{b}$ .

🎯 Exam · The n-th term of an H.P.

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

are in H.P., write

\frac{1}{a _{n}} = \frac{1}{a _{1}} + (n - 1) D

where D is the common difference of the reciprocal A.P.; then invert to get

a_{n}

. Never try to sum an H.P. directly.

★ Remember · A, G, H are in G.P.

For two numbers,

G^{2} = A H

— so A,G,H themselves form a geometric progression, and G is the geometric mean of A and H. This links the three means on the next page's inequality.

🚫 Examiner Trap · Examiner traps

(1) There is NO direct sum formula for an H.P. — invert to the reciprocal A.P., work there, then invert back. (2)

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

in H.P.

\Leftrightarrow \frac{1}{a _{1}}, \frac{1}{a _{2}}, \dots

in A.P. (3) The three means of a,b:

A = \frac{a + b}{2}

G = ab

H = \frac{2 ab}{a + b}

— and

G^{2} = A H

. (4) H is the reciprocal of the A.M. of the reciprocals, NOT

\frac{2}{a + b}

AM ≥ GM ≥ HM and Their Relations

AM ≥ GM ≥ HM: For positive numbers, the arithmetic mean is at least the geometric mean, which is at least the harmonic mean:

A \geq G \geq H

Equality holds if and only if all the numbers are equal.

🎯 Exam · The optimisation workhorse

At a

fixed sum

, the product of positive numbers is greatest when they are all equal; at a fixed product, the sum is least when equal. The simplest case is

x + \frac{1}{x} \geq 2

for

x > 0

(equality at

x = 1

★ Remember · n-number form

\frac{x _{1} + \dots + x _{n}}{n} \geq (x_{1} \dots x_{n})^{1/ n} \geq \frac{n}{\frac{1}{x _{1}} + \dots + \frac{1}{x _{n}}}

. Apply it to cleverly chosen positive quantities to bound an expression.

⚠️ Watch out · Only for positive numbers

AM-GM requires all quantities to be

positive

. Applying it to terms that can be negative (or forgetting to check the equality case for the actual maximum) is a common source of wrong bounds.

🚫 Examiner Trap · Examiner traps

(1)

A \geq G \geq H

for POSITIVE numbers only, with equality IFF all numbers are equal. (2)

G^{2} = A \cdot H

links the three means. (3) Min/Max trick: for a FIXED sum the product is greatest when all terms are equal (and vice-versa) — the engine behind most optimisation MCQs. (4) Simplest case

x + \frac{1}{x} \geq 2

for

x > 0

(equality at

x = 1

); generalises to

x^{n} + \frac{1}{x ^{n}} \geq 2

Inserting Means Between Two Numbers

★ Remember · n arithmetic means between a and b

There are

n + 2

terms in all, and b is the

(n + 2)

-th:

b = a + (n + 1) d

, so

d = \frac{b - a}{n + 1}

. The sum of the n inserted means is

n \cdot \frac{a + b}{2} = n A

(each equals the single A.M. on average).

🎯 Exam · n geometric means between a and b

Now

b = a r^{n + 1}

, so

r = (\frac{b}{a})^{1/ (n + 1)}

. The product of the n inserted G.M.s is

(ab)^{n /2} = G^{n}

(each equals the single G.M. on average).

Key facts

▸A single A.M. of a,b is $\frac{a + b}{2}$ ; a single G.M. is $ab$ (the $n = 1$ cases).
▸Count carefully: n inserted means $\Rightarrow$ $n + 2$ total terms.
▸Sum of n A.M.s $= n \times$ (single A.M.); product of n G.M.s $=$ (single G.M. $)^{n}$ .
▸For G.M.s, a and b must have the $same sign$ so the ratio is real.

⚠️ Watch out · n+1 in the denominator, not n

Because there are

n + 1

gaps between the

n + 2

terms, the step is divided by

n + 1

d = \frac{b - a}{n + 1}

r = (b / a)^{1/ (n + 1)}

. Using n is the classic error.

🚫 Examiner Trap · Examiner traps

(1) Inserting n means makes

n + 2

terms total, so b is the

(n + 2)

-th term — that fixes

d = \frac{b - a}{n + 1}

(denominator

n + 1

, not n) and

r = (\frac{b}{a})^{1/ (n + 1)}

. (2) Sum of the n A.M.s

= n \cdot \frac{a + b}{2} = n A

; product of the n G.M.s

= (ab)^{n /2} = G^{n}

. (3) Single A.M.

= \frac{a + b}{2}

, single G.M.

= ab

(

n = 1

). (4) G.M. insertion needs a,b same sign.

Sum to n Terms of Special Series

Sum to

n

terms

Series	Sum
$\sum k$	$\frac{n ( n + 1 )}{2}$
$\sum k^{2}$	$\frac{n ( n + 1 ) ( 2 n + 1 )}{6}$
$\sum k^{3}$	$[\frac{n ( n + 1 )}{2}]^{2} = (\sum k)^{2}$

★ Remember · The three to memorise

\sum_{k = 1}^{n} k = \frac{n ( n + 1 )}{2}

\sum_{k = 1}^{n} k^{2} = \frac{n ( n + 1 ) ( 2 n + 1 )}{6}

\sum_{k = 1}^{n} k^{3} = [\frac{n ( n + 1 )}{2}]^{2} = (\sum k)^{2}

Finding any sum

▸First find the n-th term $T_{n}$ of the series, then sum it.
▸If $T_{n}$ is a polynomial in n, expand using $\sum k, \sum k^{2}, \sum k^{3}$ .
▸If $T_{n}$ is a product of consecutive terms, look for telescoping.
▸ $\frac{1}{1 \cdot 2} + \frac{1}{2 \cdot 3} + \dots + \frac{1}{n ( n + 1 )} = 1 - \frac{1}{n + 1}$ .

🎯 Exam · Method of differences (telescoping)

If each term is a difference

f (k) - f (k + 1)

, the sum collapses to

f (1) - f (n + 1)

. Get there with partial fractions, e.g.

\frac{1}{k ( k + 1 )} = \frac{1}{k} - \frac{1}{k + 1}

\frac{1}{( 2 k - 1 ) ( 2 k + 1 )} = \frac{1}{2} (\frac{1}{2 k - 1} - \frac{1}{2 k + 1})

★ Remember · Σk³ = (Σk)²

The sum of the first

n

cubes is the square of the sum of the first

n

natural numbers — a neat identity worth recognising on sight.

🚫 Examiner Trap · Examiner traps

(1) Memorise the three:

\sum k = \frac{n ( n + 1 )}{2}

\sum k^{2} = \frac{n ( n + 1 ) ( 2 n + 1 )}{6}

\sum k^{3} = [\frac{n ( n + 1 )}{2}]^{2} = (\sum k)^{2}

. (2)

\sum k^{3} = (\sum k)^{2}

— a favourite identity. (3) For a polynomial-in-k term, split into

\sum k, \sum k^{2}, \sum k^{3}

and add. (4) Telescoping: write each term as a DIFFERENCE so the middle cancels (e.g.

\frac{1}{k ( k + 1 )} = \frac{1}{k} - \frac{1}{k + 1}

Arithmetico-Geometric Progression & Summation

Arithmetico-geometric progression (AGP): An A.P. multiplied term-by-term by a G.P.: the k-th term is

[a + (k - 1) d] r^{k - 1}

(for example

1 + 2 x + 3 x^{2} + 4 x^{3} + \dots

). It is summed by the

S - rS

method.

★ Remember · The S − rS trick

Write S, multiply through by the ratio r to get rS, then subtract: the staggered terms turn the middle into a plain G.P. For

∣ r ∣ < 1

the sum to infinity is

S_{\infty} = \frac{a}{1 - r} + \frac{d r}{( 1 - r ) ^{2}}

Strategy for an unfamiliar series

▸Always find the n-th term $T_{n}$ first, then sum: $S_{n} = \sum_{k} T_{k}$ .
▸Classify it: A.P.? G.P.? AGP? telescoping? a mix of $\sum k^{p}$ ?
▸Group terms, or use partial fractions, to reveal the structure.
▸Check your formula on small $n$ (e.g. $n = 1, 2$ ) before trusting it.

🎯 Exam · Recognise the building blocks

Most 'sum the series' questions are one of: a pure A.P./G.P., an AGP (

S - r S

), a telescoping sum (partial fractions), or a polynomial in k summed via

\sum k, \sum k^{2}, \sum k^{3}

. Spotting which one is half the battle.

⚠️ Watch out · Don't force a single formula

A series can be a

combination

(e.g. an A.P. plus a G.P., or

k^{2} + 2^{k}

). Split it into parts you know how to sum, rather than hunting for one closed form for the whole thing.

🚫 Examiner Trap · Examiner traps

(1) AGP

=

A.P.

\times

G.P.; sum it by the

S - r S

trick (multiply S by r and subtract — the middle collapses to a G.P.). (2) AGP sum to infinity (

∣ r ∣ < 1

S_{\infty} = \frac{a}{1 - r} + \frac{d r}{( 1 - r ) ^{2}}

. (3) For an unknown series, find the n-th term

T_{n}

FIRST, then sum

\sum T_{k}

. (4) Recognise the pattern (A.P./G.P./AGP/telescoping/

\sum k^{p}

) before grinding.

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Frequently Asked Questions

What are the most important Sequence and Series formulas for JEE Main?

This Sequence and Series formula sheet covers all the high-yield Mathematics formulas, definitions and theorems you need for JEE Main, across Arithmetic and Geometric progressions, Insertion of arithmetic and geometric means between two given numbers, Relation between A.M and G.M — each shown with the key result and, where useful, a worked example.

Is this Sequence and Series formula sheet free?

Yes — the full chapter formula sheet is free to read online, no login or payment required.

How should I revise Sequence and Series formulas?

Blurt the Sequence and Series formulas from memory, then check against this sheet to find your gaps — and practise a few previous-year questions on the chapter to make sure you can apply them under time pressure.

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