JEE Mains Maths · Sequences and Series

Arithmetic Progressions: Terms and Sums

An arithmetic progression adds the same common difference d at every step: find a term from two others, a sum from two others, and use symmetric terms and odd–even parts to shorten the algebra.

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Why this matters

Forty-four PYQs, the largest page in the chapter. Half of them are about sums, usually two given and a third asked; the rest give terms, or split the AP into odd-placed and even-placed parts. Every one comes down to two unknowns, the first term and d.

Concept 1 of 3: The nth term and the common difference

Two terms fix an AP. Between

a_p

and

a_q

there are

q-p

steps, so

d=\frac{a_q-a_p}{q-p}

, and then any term follows. Inserting

n

arithmetic means between

a

and

b

makes

n+2

terms, so

n+1

steps. Counting from the end of an AP with last term

l

, the

k

th term from the end is

l-(k-1)d

Definition

$a_n=a+(n-1)d$ .
From two terms: $d=\frac{a_q-a_p}{q-p}$ .
$n$ means between $a$ and $b$ : $d=\frac{b-a}{n+1}$ .
$k$ th term from the end: $l-(k-1)d$ .
Number of terms in $a,\dots,l$ : $\frac{l-a}{d}+1$ .

nth term

a_n=a+(n-1)d,\qquad d=\frac{a_q-a_p}{q-p}

Worked example

In an AP,

a_7=30

and

a_{15}=62

. Find

a_1

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2026 · 28 Jan 2026 Shift 1 · Q67Moderate

Example 1 · Sequences and Series · AP: Terms and Sums

The common difference of the A.P.:

a_{1},a_{2},\ldots.,a_{m}

is 13 more than the common difference of the A.P.:

b_{1},b_{2},\ldots,b_{n}

. If

b_{31}= - 277,b_{43}= - 385

and

a_{18}=

327 , then

a_{1}

is equal to

Count steps, not terms

From

a_p

a_q

there are

q-p

steps. With

n

means inserted there are

n+1

steps, not

n

Drill 10 more on the nth term and the common difference

Concept 2 of 3: Sum of n terms, and an AP given by its sum

S_n=\frac n2\big(2a+(n-1)d\big)=\frac n2(a+l)

. Two given sums make two linear equations in

a

and

d

; solve them and any third sum follows. When the sum is given as a formula in

n

, the terms come from

a_n=S_n-S_{n-1}

. An AP's sum is always of the form

An^2+Bn

, with no constant, and then

d=2A

Definition

$S_n=\frac n2\big(2a+(n-1)d\big)=\frac n2(\text{first}+\text{last})$ .
Two sums given: divide each by $\frac n2$ to get $2a+(n-1)d$ , then subtract.
Sum given as a formula: $a_n=S_n-S_{n-1}$ for $n\ge2$ , $a_1=S_1$ .
$S_n=An^2+Bn$ means an AP with $d=2A$ , $a_1=A+B$ .

Sum of n terms

S_n=\frac n2\big[2a+(n-1)d\big]

Worked example

S_5=40

and

S_{10}=155

. Find

S_{20}

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2025 · 24 Jan 2025 · Q134Moderate

Example 2 · Sequences and Series · AP: Terms and Sums

In an arithmetic progression, if

S_{40}= 1030

and

S_{12}= 57

, then

S_{30}-S_{10}

is equal to:

A constant term in S_n

If the given

S_n

has a constant term,

S_1

does not follow the pattern of

S_n-S_{n-1}

: the sequence is an AP only from the second term.

Drill 21 more on sum of n terms, and an ap given by its sum

Concept 3 of 3: Symmetric terms, equal pairs and odd–even parts

Choose unknown terms symmetrically: three as

a-d,a,a+d

, four as

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

. Then the sum fixes

a

at once. Terms the same distance from the two ends add to the same total. In an AP of

2k

terms each even-placed term is

d

more than the odd-placed term before it, so the two parts differ by

kd

. A product such as

a_1a_4

written in

d

is a polynomial: find its maximum or minimum by calculus or by completing the square.

Definition

Three terms: $a-d,a,a+d$ (sum $3a$ ). Four terms: $a-3d,a-d,a+d,a+3d$ (common difference $2d$ ).
$a_k+a_{n+1-k}=a_1+a_n$ for every $k$ .
$2k$ terms: (even-placed sum) $-$ (odd-placed sum) $=kd$ .
Each part is itself an AP with difference $2d$ .

Odd–even parts of 2k terms

S_{\text{even}}-S_{\text{odd}}=kd,\qquad a_k+a_{n+1-k}=a_1+a_n

Worked example

Three numbers in AP have sum 21 and product 231. Find them.

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2025 · 22 Jan 2025 · Q130Moderate

Example 3 · Sequences and Series · AP: Terms and Sums

Suppose that the number of terms in an A.P. is 2 k ,

k \in N

. If the sum of all odd terms of the A.P. is 40 , the sum of all even terms is 55 and the last term of the A.P. exceeds the first term by 27 , then k is equal to

Odd-placed, not odd-valued

The odd terms of an AP are

a_1,a_3,a_5,\dots

, the terms in odd positions. Their values need not be odd numbers.

Drill 10 more on symmetric terms, equal pairs and odd–even parts

Summary — formulas & gotchas at a glance

A revision cheat-sheet for the formulas and gotchas above. Click any concept name to jump back to its full explanation.

Formulas (3)

The nth term and the common difference
nth term
$a_n=a+(n-1)d,\qquad d=\frac{a_q-a_p}{q-p}$
Sum of n terms, and an AP given by its sum
Sum of n terms
$S_n=\frac n2\big[2a+(n-1)d\big]$
Symmetric terms, equal pairs and odd–even parts
Odd–even parts of 2k terms
$S_{\text{even}}-S_{\text{odd}}=kd,\qquad a_k+a_{n+1-k}=a_1+a_n$

Watch out for (3)

Count steps, not terms
→ The nth term and the common difference
A constant term in S_n
→ Sum of n terms, and an AP given by its sum
Odd-placed, not odd-valued
→ Symmetric terms, equal pairs and odd–even parts

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